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Brain&Force
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[*] posted on 24-9-2014 at 21:37
Isotopic separation at home - can it be done?


I know a NurdRage style "Crushing your expectations" disclaimer goes here: it'll probably never be practical to separate isotopes for use in the lab, and this is mostly for bragging rights.

But can isotopic separation or even enrichment be done at the amateur level? Protium/deuterium is probably the simplest one, and a process already exists to separate them: the Girdler sulfide process.

Centrifuging sounds easy to implement, but even if it actually is, it would probably be ineffective.

What are your thoughts?




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[*] posted on 25-9-2014 at 00:21


I would think the amount of time, energy, and money would make isotopic seperation unfesible for all but the most affluent of amatuer chemists. Dueterium would probably be easiest because of its realitive abundance.
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[*] posted on 25-9-2014 at 03:21


It has come up several times before on this forum, although as far as I remember nobody reported any actual experiments in this area.

For example
https://www.sciencemadness.org/whisper/viewthread.php?tid=23...
https://www.sciencemadness.org/whisper/viewthread.php?tid=22...

But I am sure you will find more threads if you search.

[Edited on 25-9-2014 by phlogiston]




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[*] posted on 25-9-2014 at 11:42


Heavy water was originally made by repeated fractional distillation.
There are of course better processes but in theory deuterium and tritium can be isolated that way. My gut feeling is this is way too slow unless you are in the middle of a war and have practically unlimited resources.
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[*] posted on 25-9-2014 at 12:19


Quote: Originally posted by Brain&Force  
I know a NurdRage style "Crushing your expectations" disclaimer goes here: it'll probably never be practical to separate isotopes for use in the lab, and this is mostly for bragging rights.

But can isotopic separation or even enrichment be done at the amateur level? Protium/deuterium is probably the simplest one, and a process already exists to separate them: the Girdler sulfide process.

Centrifuging sounds easy to implement, but even if it actually is, it would probably be ineffective.

What are your thoughts?


Centrifuge separation factors are a sensitive function of peripheral rotor speed - high sub-sonic or supersonic velocities (relative to air, but in a vacuum) are necessary to play this game. It can certainly be done, but building one rotor of non-miserable performance would be an impressive garage engineering feat.

Here is an interesting point about gas centrifuge enrichment plants for uranium is that since a large number of rotors are needed they must be cheap to build in mass production. The first generation of Dutch gas centrifuge could do 0.5 SWU/year. The more advanced P2 used by Iran now can do 5-6 SWU/yr. Really advanced carbon fiber rotors can do >100 SWU/yr these days. But the market price for one SWU (separative work unit) is only $89. So if a rotor has a service life of 4 years, and must turn a profit in that time, a 5 SWU unit would need to cost <$1500.

Isotope separation, in macroscopic amounts, with substantial degrees of separation is very hard to do, before WWII separated isotope products were in miniscule supply. D2O was available only because of surplus electricity at the new Norsk Hydro dams.

If you want to demonstrate or lay claim to isotope separation, your best bet is to build a simple ion mass spectrometer, along the lines of Aston. Using a phosphor screen and a magnetic field you can show a beam of ions splitting into two different masses.

An artificially prepared mixture of deuterium and light hydrogen would get you two equally strong beams, with a large mass ratio.

Otherwise, consider using neon which has two major isotopes (20 and 22) which are present in a 10:1 ratio. Neon was the first element to be separated by isotopes.
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[*] posted on 26-9-2014 at 00:22


Sounds like a proper challenge :)

I believe the first enrichment of deuterium was performed by distillation of hydrogen, boiling about 5 liters of liquid hydrogen down to 1 ml, enriching the deuterium by a factor of about 7.
The results were confusing at first, because the hydrogen had been prepared via electrolysis, which depleted it of deuterium. The enrichment process more or less cancelled the depletion from the distillation.




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[*] posted on 26-9-2014 at 07:08


Quote: Originally posted by phlogiston  
Sounds like a proper challenge :)

I believe the first enrichment of deuterium was performed by distillation of hydrogen, boiling about 5 liters of liquid hydrogen down to 1 ml, enriching the deuterium by a factor of about 7.
The results were confusing at first, because the hydrogen had been prepared via electrolysis, which depleted it of deuterium. The enrichment process more or less cancelled the depletion from the distillation.


That's a pretty good concentration increase for one stage, and simple, but at the cost of throwing away nearly all the D in the feed (~99.9% of it). The separation factor for LH2 distillation (1.5) is not especially large, for H2O-H2S exchange it is 1.8-2.3.

Water electrolysis separation factor is 5-10, but the energy cost is enormous. With electrolysis you can do the same trick as with LH2 distillation in which you electrolyze are starting batch down to a very small fraction, and get a substantial enrichment ratio, but at the cost of throwing away nearly all the deuterium. The water is cheap, the electrolysis is not.

A nice description of a laboratory apparatus for deuterium and tritium concentration with electrolysis (and its operation) is found in "The Electrolytic Enrichment Of Tritium And Deuterium For Natural Tritum Measurements" by Ostlund and Werner is described here:
http://www.iaea.org/inis/collection/NCLCollectionStore/_Publ...

Usually lab concentration of D by electrolysis (the purpose is for analysis, not high purity isotope) is done in ~5 stages, with 10:1 volume reduction in each stage. So to get 1 mL of enriched product, something like 100 L of water must be electrolyzed. At 80 kwh/kg water (50% energy efficiency) for electrolysis that is 8000 kwh. At the usually quoted residential electricity rate of ~$0.10/kwh, this is $800 (some people pay much more for their marginal energy consumption). You can buy one kg of high purity heavy water from United Nuclear for this price.

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[*] posted on 26-9-2014 at 07:35


"Isotopic separation at home - can it be done?"
You already are doing it to some degree
http://en.wikipedia.org/wiki/Isotopic_signature

Now, can someone home-build the kit needed to measure the changes in isotopic ratios?

Incidentally, while the difference in properties between H and D is bigger than for any other pair of isotopes, the absolute concentration of D is very low so you need to process a lot of water to get a small amount of D2O.
It might be easier to demonstrate some sort of separation with other isotope pairs.


[Edited on 26-9-14 by unionised]
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[*] posted on 26-9-2014 at 20:40
Ion Gun


I think your best bet for getting good separations of significant amounts of natural isotopes is the same way Oak Ridge does it: ion gun magnetic field separation.

The separation calutrons could operate with beam currents at least up to 46 mA:
http://www.osti.gov/scitech/servlets/purl/4246415

At that current one mole of ions goes through the beam every 24 days. I don't know what beam currents an amateur rig might do, but surely in the milliampere range is possible.

Deuterium, as others have pointed out, is a bad candidate to separate since it is so scarce. For every mole processed there is very little product. A separation candidate I might suggest is lithium-6 and lithium-7. They have a large mass ratio, and lithium-6 is not extremely rare: 7.42% of natural lithium.

Neon is also a good candidate (as I noted previously) particularly for an ion gun. A noble gas gun is probably a bit simpler than a lithium gun.

If you stick with relatively abundant high mass ratio isotopes you will have it much easier than the Oak Ridge calutrons that were called on, for example, to separate the isotopes of tin of which there are ten, with some pairs separately by only 0.85% in mass (even uranium had a 1.3% ratio).

(This is really the same as my Aston mass spectrometer suggestion, but on steroids and crank.)
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[*] posted on 10-11-2014 at 12:49


Does anyone know the magnetic field strength needed to run a Calutron?

[Edited on 10-11-2014 by blogfast25]




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[*] posted on 10-11-2014 at 13:27


Quote: Originally posted by blogfast25  
Does anyone know the magnetic field strength needed to run a Calutron?

[Edited on 10-11-2014 by blogfast25]


"a calutron, which is an electromagnetic device for the separation of isotopes of polyistopoic material and which employs powerful electromagnets setting up magnetic fluxes which may reach the order of 4000 gauss."

http://www.google.com/patents/US2628297

More usefully though, you need to look up the physics of the Calutron and work with the numbers to see if what you can plausibly implement. The basic physics of the EM separation process is not very complex.

You are not going to be able to clone an Oak Ridge calutron, design compromises will be needed.
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[*] posted on 10-11-2014 at 17:35


The magnetic field strength needed varies directly with the beam energy and indirectly with the turn radius. so by tuning the other two factors a home scale calutron should be perfectly feasible (No promises on the beam current though).



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[*] posted on 11-11-2014 at 08:17


Quote: Originally posted by Arcuritech  
The magnetic field strength needed varies directly with the beam energy and indirectly with the turn radius. so by tuning the other two factors a home scale calutron should be perfectly feasible (No promises on the beam current though).


Right - as I noted above you are just making a type of ion mass spectrograph. J.J. Thomson accomplished this with his ion beam experiments in 1907, and which Aston developed into a laboratory instrument in 1909. They weren't working with huge Calutron magnets.
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[*] posted on 17-3-2016 at 07:34


You can get isotope separation through a number of simple chemical operations. For example, crystals usually preferentially incorporate isotopically light ions as they grow. The difference is usually slight, e.g., the 44Ca/40Ca ratio in CaCO3 grown at room temperature is about 1.0‰ (0.1%) lower than the parent solution. However, you can use rayleigh distillation to leverage that small shift into a larger one. Because the isotopic composition of the growing surface is offset from that of the solution by a constant amount (the fractionation factor), in a closed system with a diminishing supply of dissolved material as the crystal grows the material remaining in solution will become very heavy-- with a fractionation factor of 1.0‰ you will get a heavy isotope enrichment of about 1.4% in the last 1% of the solution. Of course, as you are losing 99% of your starting material, getting higher enrichments quickly becomes a problem (as it does with all enrichment processes).

You can get higher enrichments with lighter elements, not because they are lighter, but because the relative mass differences between isotopes tends to be greater. Also, the lower the temperature the better the enrichment.

Column chemistry also fractionates isotopes, which can be a real problem in isotope chemistry. I can't recall exactly, but I know that Li is especially prone to this, and that you can get percent level changes in 7Li/6Li on standard columns. We once calculated that you could get a ca 20% enrichment on a single pass through a long (say 3 meter) column, so making fairly pure 7Li at home might not be out of the question.

Working with gas phases helps too, because in gasses you get the full effect of the kinetic differences between isotopes that drives most fractionation. In gasses diffusing across a membrane the fractionation factor will be proportional to the square root of the difference in mass between isotopes, which usually is greater than what you can get in a wet system.




[Edited on 17-3-2016 by Joe Skulan]
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[*] posted on 17-3-2016 at 08:20


Regarding separation on columns, I can confirm that isotope effects are actually commonly observed in normal LC-MS work. Deuterated compounds are often used as the internal standards, and it is not uncommon for them to experience a shift in retenention time to such a degree that they become unsuitable as an internal standard (for instance because the co-elute with some component that interferes with its ionization in the MS, which is not the case with the analyte of interest).



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[*] posted on 17-3-2016 at 08:58


That's why isotope double spikes are nice.
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[*] posted on 22-6-2016 at 21:19


I just realized that codyslab's heavy water series has yet to be mentioned. Here's a link: https://www.youtube.com/playlist?list=PLKhDkilF5o6_MTogGYYdE...

Over the course of about two months (or two years if you include how long his batteries had been electrolyzing water before he started the project) he successfully isolated about \( \frac{1}{2} \)g of heavy water at a high enough concentration to demonstrate that its ice will sink in normal water.




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[*] posted on 22-6-2016 at 21:25


Quote: Originally posted by Arcuritech  
I just realized that codyslab's heavy water series has yet to be mentioned. Here's a link: https://www.youtube.com/playlist?list=PLKhDkilF5o6_MTogGYYdE...

Over the course of about two months (or two years if you include how long his batteries had been electrolyzing water before he started the project) he successfully isolated about \( \frac{1}{2} \)g of heavy water at a high enough concentration to demonstrate that its ice will sink in normal water.

I was just thinking about this in relation to the current thread on K40. Cody has done a follow-up where he used a chemical process rather than electrolysis to enrich heavy water to reasonable levels within the space of five days. He also noted in this experiment that whatever process was used to produce the NaOH that he used, it also enriched the deuterium content.

I guess the conclusion is that isotope enrichment is certainly possible for the amateur chemist but the process is both inefficient and slow. Actual detection of your results is likely to be a bigger hurdle than enrichment in most cases.




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[*] posted on 23-6-2016 at 00:59
let's design a calutron


On the topic of building a calutron, let's design a calutron to separate lithium into its isotopes. Let's make it like the original, ions travel 180 degrees before hitting the collector cup.

The cyclotron formula:
E = q^2 B^2 r^2 / (2m)

Energy (E): let's assume the ion source is a simple ion gun, no fancy accelerator stages. The largest transformer I have is 75kV 5ma and I doubt many people can beat that, so I will assume the energy of both Li-6 and Li-7 is 75keV.

Charge (q): the charge of Li+ ions is 1 elementary charge

Magnetic field (B): iron cores saturate at 1.6 T so I will assume that will be our field, although it is highly ambitious to actually achieve this field strength

Radius (r): that is what we are solving for

Mass (m): Li-6 is 6 amu, Li-7 is 7 amu

If we plug in all the numbers and solve for radius, we get:

Li-6, 60.4 mm
Li-7, 65.2 mm

Therefore the magnetic poles will have to be a minimum of 130.4 mm (5 inches) in diameter.

Now let's take a look at a real magnet, like a Varian V-3400 NMR magnet. It is 9 inches in diameter and "The magnet weighs 1780 pounds, it requires 40 volts at 168 amps, 7Kilowatts, to produce the maximum field of 1.2 Tesla." Recalculating the above values with 1.2 tesla gives 80.5 mm and 86.9 mm for Li-6 and -7 respectively. The diameter of the ion arc is 6.842 inches which will fit nicely into a 9" magnet.

The targets for Li-6 and Li-7 will have to be 2x(86.9 - 80.5) = 12.8 mm apart. That means the spot size of the beam also has to be 12.8 mm or less in order not to overlap (though some overlap is expected anyway). Sounds generous but that means the ion beam half-angle should be no greater than 1.34 degrees, so simple optics and tuning will be required.

Now suppose a lithium calutron was built, how long would it take to process a gram (144 mmol) of lithium? That's 8.672×10^22 atoms which is 13.89 kC of charge. At 5mA it will take 32 days 3 hours 40 minutes, assuming 100% efficiency. In reality, it will be more like 1 year to process 1 gram of natural lithium metal. More current will speed things up, but even the transformer I used for this example is already massive.

In 32 days it will cost $20.16 in electricity to run the transformer (5mA * 75 kV * 32 days * [7 cents per kilowatt hour]), $376.32 for the electromagnet, and possibly a similar amount for the vacuum pumps. If it did end up taking an entire year it would cost $230 and $4292 for the two.

If you just want to prove a point, making 1 mg should be enough and will be much cheaper.

[Edited on 23-6-2016 by DoctorOfPhilosophy]

[Edited on 23-6-2016 by DoctorOfPhilosophy]
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[*] posted on 23-6-2016 at 07:26


Quote: Originally posted by DoctorOfPhilosophy  
On the topic of building a calutron, let's design a calutron to separate lithium into its isotopes. Let's make it like the original, ions travel 180 degrees before hitting the collector cup.

The cyclotron formula:
E = q^2 B^2 r^2 / (2m)

Energy (E): let's assume the ion source is a simple ion gun, no fancy accelerator stages. The largest transformer I have is 75kV 5ma and I doubt many people can beat that, so I will assume the energy of both Li-6 and Li-7 is 75keV.

Charge (q): the charge of Li+ ions is 1 elementary charge

Magnetic field (B): iron cores saturate at 1.6 T so I will assume that will be our field, although it is highly ambitious to actually achieve this field strength

Radius (r): that is what we are solving for

Mass (m): Li-6 is 6 amu, Li-7 is 7 amu

If we plug in all the numbers and solve for radius, we get:

Li-6, 60.4 mm
Li-7, 65.2 mm

Therefore the magnetic poles will have to be a minimum of 130.4 mm (5 inches) in diameter.

Now let's take a look at a real magnet, like a Varian V-3400 NMR magnet. It is 9 inches in diameter and "The magnet weighs 1780 pounds, it requires 40 volts at 168 amps, 7Kilowatts, to produce the maximum field of 1.2 Tesla." Recalculating the above values with 1.2 tesla gives 80.5 mm and 86.9 mm for Li-6 and -7 respectively. The diameter of the ion arc is 6.842 inches which will fit nicely into a 9" magnet.

The targets for Li-6 and Li-7 will have to be 2x(86.9 - 80.5) = 12.8 mm apart. That means the spot size of the beam also has to be 12.8 mm or less in order not to overlap (though some overlap is expected anyway). Sounds generous but that means the ion beam half-angle should be no greater than 1.34 degrees, so simple optics and tuning will be required.

Now suppose a lithium calutron was built, how long would it take to process a gram (144 mmol) of lithium? That's 8.672×10^22 atoms which is 13.89 kC of charge. At 5mA it will take 32 days 3 hours 40 minutes, assuming 100% efficiency. In reality, it will be more like 1 year to process 1 gram of natural lithium metal. More current will speed things up, but even the transformer I used for this example is already massive.

In 32 days it will cost $20.16 in electricity to run the transformer (5mA * 75 kV * 32 days * [7 cents per kilowatt hour]), $376.32 for the electromagnet, and possibly a similar amount for the vacuum pumps. If it did end up taking an entire year it would cost $230 and $4292 for the two.

If you just want to prove a point, making 1 mg should be enough and will be much cheaper.


Thanks. Yep that sounds like a Calutron.

If you are after only the rarer, and more interesting Li-6, the actual minimum pole diameter is only 120.8 mm/161 mm since you are throwing the Li-7 beam away. Edge cutting adjustment only would make design and operation simpler (the U-235 producing Calutrons used this fact).

I have not tried to play with the parameters in this design, but assuming you are only working on the milligram scale, can you suggest any design compromises that might make the unit a little more attractive to the (very ambitious) hobbyist?

If point proving only, perhaps even an Aston-type mass spectrometer would do. You aren't collecting material, but you can show the separation of isotopes.

Since higher beam energy increases the factor B^2r^2 (field strength and radius) which we should want to reduce, why not use a lower beam energy? What is the trade-off there? The actual Calutrons used for isotope separation at Oak Ridge often had ion source voltages of several keV to 30 keV or so, not as high as 75 keV. The original U-235 Calutrons had a 0.35 T field.

[Edited on 23-6-2016 by careysub]
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[*] posted on 23-6-2016 at 09:40


With respect to lithium isotopes the use of ion exchange resins have been suggested as well as other procedures.
See attached report.

Attachment: LithiumIsotopeSeparation.pdf (427kB)
This file has been downloaded 122 times

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[*] posted on 23-6-2016 at 11:13


It's actually very easy to enrich isotopes; even at home.
Put some water in the ice cube tray in the freezer and wait until half of it is frozen.
Pour off the liquid and then put the ice in another jar.
There we have it- isotopic enrichment.

The difficult problem is measuring the extent of the enrichment.
:-)

Seriously, if you enrich these things, how will you know?
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[*] posted on 23-6-2016 at 15:58


Quote: Originally posted by unionised  


The difficult problem is measuring the extent of the enrichment.
:-)

Seriously, if you enrich these things, how will you know?

I can already list 3 measurable effects:
1°) Change of density: since different isotopes display different mass for the same atomic volume, the density will change...
Example:
D2O is denser than H2O, it has also different physico-chemical properties like mp, boiling point, refractive index, speed of light into the media, ...
D2O ice cubes sink into water unlike conventional ice cubes...
The name "heavy water" comes from somewhere...

2°) Change of molarity, molality, concentration: since the average atomic mass changes, a given weight of enriched material will change the amount of moles...
Example:
A)
6LiOH = 23 gr/mole
while
7LiOH = 24 gr/mole

So 23g of 6LiOH is 1 mole and 1M if dissolved into 1L of demi water
But 23g of 7LiOH is 0.9583333 mole and 0.9583333 M if dissolved into 1L of demi water


So all physico-chemical properties linked will be changed like the osmotic pressure, the vapour pressure, the boiling point and the freezing point of the solution, the conductivity, ...

B) If you have D2O, it is like 50 M, while H2O is like 55.555 M

3°) If radioactive (cf KI), it will change the flow of radioactivity arround it: more or less, depending if you enrich with the most or the least
radioactive...
Example:
One gr of non-enriched material in a cylindrical Geiger counter vs one gr of enriched material will display a change in radiocativity.

[Edited on 24-6-2016 by PHILOU Zrealone]




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[*] posted on 23-6-2016 at 16:39


Quote: Originally posted by PHILOU Zrealone  
Quote: Originally posted by unionised  


The difficult problem is measuring the extent of the enrichment.
:-)

Seriously, if you enrich these things, how will you know?

I can already list 3 measurable effects:
1°) Change of density: since different isotopes display different mass for the same atomic volume, the density will change...
Example:
D2O is denser than H2O, it has also different physico-chemical properties like mp, boiling point, refractive index, speed of light into the media, ...
D2O ice cubes sink into water unlike conventional ice cubes...
The name "heavy water" comes from somewhere...

2°) Change of molarity, molality, concentration: since the average atomic mass changes, a given weight of enriched material will change the amount of moles...
Example:
A)
6LiOH = 23 gr/mole
while
7LiOH = 24 gr/mole

So 23g of 6LiOH is 1 mole and 1M if dissolved into 1L of demi water
But 23g of 7LiOH is 0.9583333 mole and 0.9583333 M if dissolved into 1L of demi water


So all physico-chemical properties linked will be changed like the osmotic pressure, the vapour pressure, the boiling point and the freezing point of the solution, the conductivity, ...

B) If you have D2O, it is like 50 M, while H2O is like 55.555 M

3°) If radioactive (cf KI), it will change the flow of radioactivity arround it: more or less, depending if you enrich with the most or the least
radioactive...
Example:
One gr of non-enriched material in a cylindrical Geiger counter vs one gr of enriched material will display a change in radiocativity.

[Edited on 24-6-2016 by PHILOU Zrealone]


These are good points but I think what unionised means is 'how would you follow enrichment progress?' E.g. the difference between the heaviest and lightest KOH is about 5%. To follow enrichment progress by titration (e.g.) of successive enrichments would be hard.

MS is what's really needed.




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[*] posted on 23-6-2016 at 20:38


Quote: Originally posted by blogfast25  
...I think what unionised means is 'how would you follow enrichment progress?' E.g. the difference between the heaviest and lightest KOH is about 5%. To follow enrichment progress by titration (e.g.) of successive enrichments would be hard.

MS is what's really needed.


Evaluation of every stage is almost certainly unnecessary.

In the case of most chemical processes, the separation factor is a known value and therefore dead reckoning can be used to replace evaluation of individual steps. Only after a significant number of iterations of concentration have passed would you bother to measure the separation.

This estimation based approach may not work as well for calutrons and other mass-spectrometry based separations, so one could preform a "calibration run" ahead of time by temporarily replacing the collector (receiver? catch? someone help me out on this) with a fused-quartz plate (or borosilicate, SiO2 and the like glow blue when bombarded with charged particles) to find the beam withs and separations as well as to confirm the calculated beam positions.

You would want to do this full beam current, because some effects like charge-induced beam spreading will only show themselves at full current.




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