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How can KCl be enriched in 40K at home?

  1. Feb 11, 2019 #1

    ORF

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    Hello,

    I have pure KCl, but its natural radioactivity is low. I would like to increase it at least by a factor of 10. Is there any simple way (diy) for enriching the natural KCl in 40K?

    Alternatively, is there a compound with higher concentration of potassium (in volume) than KCl, and that you can have at home?

    Thank you for your time.

    Regards,
    Alvaro.
     
  2. jcsd
  3. Feb 11, 2019 #2

    phyzguy

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    I think the answer is no to both questions. What exactly are you trying to do? Radioactive materials are strictly regulated, for good reasons.
     
  4. Feb 11, 2019 #3

    .Scott

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    I am not sure what the rules and regulations are.
    Here is a site which will could sell you the material - and perhaps advise you on what would be required for the handling, reporting, and other regulatory requirements:
    https://www.buyisotope.com/potassium-40_isotope.php

    Realistically, the only way you're going to get a 10-fold increase in the K40 concentration is to have someone do it for you.
     
  5. Feb 11, 2019 #4

    .Scott

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    A quick look at the hazard you're considering:

    If you procured 20 grams of KCl with K40 enriched 10-fold, it would include about 9 grams of Potassium. It would also include about 610 Becquerels of K40 - thus about 610 radioactive decays per second.

    Of course, ingesting 20 grams of KCl in a shot presents some risk just because of the chemistry.
    But perhaps the best way to assess the radiation health risk it to determine its "Banana Equivalent Dose": https://en.wikipedia.org/wiki/Banana_equivalent_dose

    The BED for 9 grams of Potassium is: 9000mg/422mg = 21 BEDs.
    Thus about:
    2.1 microsieverts of exposure
    roughly 21% of the typical daily radiation exposure.
    committed dose equivalent over 50 years: 1.7 microsieverts

    In contrast, a dental xray will expose you to between 5 and 50 microsieverts.
     
  6. Feb 11, 2019 #5

    Vanadium 50

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    40K is not that radioactive. Nevertheless, if you try and buy some, you're going to have to deal with all the regulations.

    I don't think you can do this with chemistry. 40K is chemically identical to other isotopes, and even if there were a process, you would need to separate it from both 39K, which is lighter, and 41K, which is heavier.

    If you buy it, I expect a factor of 10 enrichment would cost $5,000-10,000 per gram.
     
  7. Feb 11, 2019 #6

    Ygggdrasil

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    If you do find a DIY method for isotope enrichment at home, I'm sure there are plenty of people in Iran and other countries who would be interested in hearing about it.
     
  8. Feb 11, 2019 #7

    TeethWhitener

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    In principle, you could separate out the isotopes by their reaction rates in some appropriately chosen reaction:
    https://en.m.wikipedia.org/wiki/Kinetic_isotope_effect

    In practice, this will be very difficult for potassium, for two reasons:
    1) The kinetic isotope effect tends to be a function of the mass ratio between isotopes, as it is fundamentally a result of different zero point vibrational energies. So it works well for separating hydrogen/deuterium, but not so much for heavier atoms.

    2) Potassium compounds tend to be either really unreactive (like KCl) or really reactive (like potassium cyclopentadienide). I suppose you may be able to figure out some really clever salt metathesis, but it would probably be really inefficient.
     
  9. Feb 11, 2019 #8

    TeethWhitener

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    To answer your second question, KF has a slightly higher volume fraction of potassium than KCl. It’s not particularly safe, though. KH has a much higher volume fraction of potassium, but it’s even less safe. At that point, you might as well be working with potassium metal directly.
     
  10. Feb 12, 2019 #9

    DrDu

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    I wonder whether some enrichment can be obtained by ion exchange chromatography.
     
  11. Feb 12, 2019 #10

    TeethWhitener

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    In principle, as long as there is a contribution from interatomic vibrations and the zero point energy differences of the reactant states are different from the zero point energy differences of the transition state resonances, any chemical interaction would work. For instance, pure D2O boils at a slightly higher temperature than pure H2O, and in fact, distillation is used industrially to enrich heavy water.

    Would it work for potassium? I doubt it would be very efficient, based on reason #1 from post #7.
     
  12. Feb 12, 2019 #11

    DrDu

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  13. Feb 12, 2019 #12

    Vanadium 50

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    Again, you can't do this with chemistry. If you use any of the various techniques to favor heavy nuclei (the bulk of potassium is 39K), you will build up the 40K component, but you will also build up the 41K component even faster. The best you can possibly do is a factor of 2.

    This assumes your reagents are isotopically pure. If you are using anything water-based, you will quickly discover the process - whatever it is - is doing a much better job of getting the deuterium out of the water (and much, much better at getting the 18O out of the water) than the 40K out of the potassium.

    Of course, you can always use a mass spec. That's essentially how they make it commercially. A good rule of thumb is that the cheapest isotopes you can get by this process run a few thousand per gram.
     
  14. Feb 12, 2019 #13

    TeethWhitener

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    There might be a clever way to integrate or hybridize this with a recycling HPLC to get a functionally unlimited column length. We used to use one to separate out C60 from He@C60. It takes forever, wastes a bunch of solvent, and is extremely energy intensive and inefficient. Existing isotope separation methods are probably cheaper. Nonetheless, I don't think it's out of the realm of possibility. But definitely not a DIY at home project.

    Or with a factor of 1.007 enrichment on each pass, all you'd need to do is run the column ##\frac{\ln 10}{\ln 1.007} \approx 330## times to get a factor of 10 enrichment. o0)

    Edit: just looked at the paper: it's looking specifically at 41K/39K ratios. It doesn't say what the 40K enrichment is. So just run the column a few more thousand times and take the middle fraction.
     
  15. Feb 12, 2019 #14

    phyzguy

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    How could this work, given Vanadium50's points in post #12?
     
  16. Feb 12, 2019 #15

    TeethWhitener

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    The point is that the three isotopes all have different retention times (different partition coefficients, if you want to think of it that way). So given a long enough column, the three isotopes would separate into different fractions.

    Edit: just so we're clear. It's completely impractical. The effective column length would probably have to be hundreds of kilometers. Even to separate out 41K would take a column length of 330×4 = 1320 meters.
     
  17. Feb 12, 2019 #16

    DrDu

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    most methods for isotopic enrichment are either impractical or absurdly energy intensive.
     
  18. Feb 12, 2019 #17

    Vanadium 50

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    If you need something isotopically pure, it's best if you can use an element that is already a single isotope. Thta's why the LHC's lead is much more expensive than RHIC's gold: gold is isotopically pure when you dig it out of the ground. That's why aluminum good, cobalt good, but tin? Very very bad.
     
  19. Feb 12, 2019 #18

    Vanadium 50

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    I have been so focused on the potassium, I forgot about the chlorine.

    You can't use KCl. There's a thousand times as much 37Cl as 40K. You'll need another compound or you will find yourself separating the chlorine and not the potassium.
     
  20. Feb 12, 2019 #19

    TeethWhitener

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    The relevant interaction in ion exchange chromatography is electrostatic, where the analyte of interest and the stationary phase have opposite charges. So in the case of (e.g.) cation exchange chromatography, separation only happens between positively charged species:
    https://en.m.wikipedia.org/wiki/Ion_chromatography
     
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