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Author: Subject: Deuterium!
elementcollector1
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biggrin.gif posted on 17-3-2013 at 20:43
Deuterium!


A friend of mine (SWIM? Ha ha.) is looking to build a device to generate him some deuterium. Now, the current idea is to electrolyze massive amounts of water and separate the deuterium from the hydrogen. I think this isn't too good of an idea, because the scale of this machine would have to be huge to get any significant quantity of deuterium. So, my idea is to irradiate hydrogen or water with neutrons from an alpha source (Am-241) and a neutron emittor (Be metal). I don't think this would work, as this seems almost too easy. But whose method is better, and why?

(This is in Beginnings because I basically am speculating from my armchair. No, my chair does not have armrests.)




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[*] posted on 17-3-2013 at 23:44


I think that in a home setting it simply is impossible to make deuterium. If you have an alpha-source (like Am-241), then how much do you have? A few micrograms? At most a few nanomole of atoms. From these atoms you may be capable of making a few nanomole of neutrons and you may be capable of directing a few percent of these neutrons into your reaction vessel. So, even if the reaction is 100% efficient, you may obtain a few hundreds of picomoles of deuterium atoms. Even in a ml of plain tap water you already have more deuterium atoms (IIRC appr. 0.02% of all hydrogen atoms are deuterium atoms).

The only viable way is separation by means of electrolysis. But you need to do the electrolysis hundreds or even thousands of time, starting with tonnes of material, and ending with a few hundreds of ml of heavy water. At an industrial scale this is possible and it is possible to make deuterium at a cost of a few euros (dollars) per gram, but in a home setting it is impossible to do this.




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phlogiston
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[*] posted on 18-3-2013 at 00:51


There are other methods than electrolysis that are faster.
To make small quantities in an amateur scientist setting, you should focus on methods that give the most enrichment per step, and not so much on energy efficiency. Some chemical reactions have a strong isotope effect.
Look up the Girdler sulfide (GS) process for instance, but there are probably other ways.

[Edited on 18-3-2013 by phlogiston]




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elementcollector1
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[*] posted on 18-3-2013 at 07:15


Girdler sulfide method sounds interesting, but Wiki states that it only enriches deuterium content up to 20%, and the rest of the stuff has to be purified by vacuum distillation.



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[*] posted on 18-3-2013 at 12:26


I would add to Woelen's comment the fact that roughly one neutron is produced for every 30 million Alpha's making the result even worse. I built a few generators using smoke detector Am and Be foil. One day I got tired of looking through a massive pile of boxes as I stored them years ago and I thought maybe I could find the box with my Neutron detector. Ended up finding them the hard way after that idea failed. They just do not emit all that many Neutrons and I could not see the difference between them and the random false pulses my BF3 tubes produce from Gamma's.

Elementcollector1 "A friend of mine (SWIM? Ha ha.) is looking to build a device to generate him some deuterium."

This just hit me. Damn that really is funny!


[Edited on 3-19-2013 by IrC]




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elementcollector1
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[*] posted on 19-3-2013 at 07:12


As I understand it, the H2S (carrying deuterium) runs from the pressurized, hot water solution to the cold solution. What if I used a lot of hot water, and only a small amount of cold water: Would this allow higher concentrations, or have no effect? I think it would allow for higher concentrations.



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watson.fawkes
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[*] posted on 19-3-2013 at 08:23


There's a recent thread on deuterium enrichment: https://www.sciencemadness.org/whisper/viewthread.php?tid=22897.
Quote: Originally posted by elementcollector1  
Girdler sulfide method sounds interesting, but Wiki states that it only enriches deuterium content up to 20%, and the rest of the stuff has to be purified by vacuum distillation.
That's not what it says.
Quote:
Normally in this process, water is enriched to 15–20% deuterium. Further enrichment to "reactor-grade" heavy water (> 99% deuterium) is done in a vacuum distillation unit.
The second stage doesn't have to be vacuum distillation, rather that it's normally in two stages. The reason is energy efficiency. Since the boiling point of heavy water (and semi-heavy DHO) is higher than that of water, if you try distilling raw feedstock, you would have to boil away all the light water to get any enrichment at all. The GS process is more efficient than that. Separation by distillation, though, can be effective when there's enough deuterium content. At some point there's a tradeoff and you switch enrichment processes.
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elementcollector1
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[*] posted on 19-3-2013 at 09:03


Oh, I see. Still, according to the Wiki entry, the feedstock ratio for the GS process is 340 water:1 D2O product. So, if I put in say, 1 gallon of water, I would get out (assuming they're right, which is a big assumption) 16.25mL of pure D2O. That's actually a fairly good ratio in the sense of things, but I doubt it'll be the correct one for the home experimenter.
So, couldn't I just continue running the GS process with the same H2S and a continual input of water until I had pure D2O? What cutoff exists in terms of efficiency for the sulfide process (I'm guessing energy consumption)?




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[*] posted on 20-3-2013 at 11:19


Quote: Originally posted by elementcollector1  
Oh, I see. Still, according to the Wiki entry, the feedstock ratio for the GS process is 340 water:1 D2O product. So, if I put in say, 1 gallon of water, I would get out (assuming they're right, which is a big assumption) 16.25mL of pure D2O. That's actually a fairly good ratio in the sense of things, but I doubt it'll be the correct one for the home experimenter.
So, couldn't I just continue running the GS process with the same H2S and a continual input of water until I had pure D2O? What cutoff exists in terms of efficiency for the sulfide process (I'm guessing energy consumption)?


How did you get 16.25 mL?
1 gallon = 3785.411865 mL

16.25mL / 3785.411865 mL = 0.429 %
but deuterium is 0.015% of natural hydrogen.

On top of that, some of the water is DHO. I think I read that if you add 5 mL H2O to 5 mL D2O, 50% ends up as DHO, 25% as H2O, 25% as D2O.
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watson.fawkes
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[*] posted on 20-3-2013 at 14:28


Quote: Originally posted by elementcollector1  
[...] according to the Wiki entry, the feedstock ratio for the GS process is 340 water:1 D2O product.
[...]
So, couldn't I just continue running the GS process with the same H2S and a continual input of water until I had pure D2O? What cutoff exists in terms of efficiency for the sulfide process (I'm guessing energy consumption)?
Read again. That ratio is 340,000 : 1. At 155 ppm D-H fraction, the best you could possibly do is 6450 : 1. So it's only about a factor of 50 less feedstock-efficient, but since the feedstock is water, it doesn't matter. The energy input is what drives the economics.

I don't have a handy reference for the detailed efficiency of the GS process. On the other hand, isotopic exchange processes generate first-order differential equations (if I understand the dynamics correctly). So the modelling isn't that hard. Dig into it if you care enough.
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[*] posted on 1-5-2015 at 12:43


A guy called Cody seems to have succeeded at this challenge and shared his efforts on youtube in a series of 5 video's. Since the subject has come up a couple of times, perhaps some of you might find it interesting to watch:

https://www.youtube.com/watch?v=VC4dk1JU4tQ




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[*] posted on 2-5-2015 at 06:12


Quote: Originally posted by IrC  
I would add to Woelen's comment the fact that roughly one neutron is produced for every 30 million Alpha's making the result even worse. I built a few generators using smoke detector Am and Be foil. One day I got tired of looking through a massive pile of boxes as I stored them years ago and I thought maybe I could find the box with my Neutron detector. Ended up finding them the hard way after that idea failed. They just do not emit all that many Neutrons and I could not see the difference between them and the random false pulses my BF3 tubes produce from Gamma's.
[Edited on 3-19-2013 by IrC]


I`ve seen different variation on the Am/Be source depending on geometry but its pretty much on point.
did the same experiment with the same equipment ! I second the result with the BF3 detector .... nothing much to speak of










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[*] posted on 2-5-2015 at 15:38


Hydrogen has a very low rate of neutron capture. Deuterium enrichment is simply not efficient enough for useful yields.
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[*] posted on 2-5-2015 at 16:45


heavy water is not that expensive! an electrolysis shouldn't be impossible...even for a cheap home scientist...
if you really want deuterium it should not be a problem ....




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[*] posted on 3-5-2015 at 09:59


Quote: Originally posted by IrC  
roughly one neutron is produced for every 30 million Alpha's

Roughly 1 per 30'000 I think or around 1 neutron per second from 1uCi of alpha source.

The conclusion still follows.
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Polverone
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