Archive for the ‘Isotope Separation’ Category

Enrichment: Visualizing the Value Function

Tuesday, August 24th, 2010


V(x) = (1 – 2x)*ln((1 – x)/x)

I’m a math minor, and an equation like that is not a very appealing way to begin a blog post. But I got the unpleasant part out of the way early because the good news that this equation (the value equation) is totally central towards understanding how much separative work we need to separate isotopes from one another, and the even better news is that the value function makes a lot more intuitive sense than we might initially think.

Let me show you what the value function looks like when you plot it on a graph. The input value “x” can only range from zero to one. No negative numbers or values greater than one allowed here!


You can see that the value function looks like a deep valley with very steep sides. Reading the values of the graph, you can see that if the input “x” is one-half (0.5) then the value function is zero. And it’s hard to see, but you can imagine that if the value function is zero or one, then it has an infinite value. So what does all this actually mean?

To help answer this question recently I bought a bunch of red and white marbles online (www.megaglass.com if you’re interested). Red and white marbles are going to represent two different nuclear isotopes in my explanation of the value function. They’re pretty good at doing this because they’re the same size and only differ in color.

Let’s say I only have white marbles in some kind of mixture. Well, then it isn’t much of a mixture! But according to the value function, when I have all white marbles in my mixture (and we’ll say that “x” is the fraction of red marbles in my mixture) then my value function would be:


V(0) = (1 – 2*0)*ln((1 – 0)/0) = 1*ln(1/0) = infinite


Let’s imagine that I only have red marbles in my “mixture” (x = 1.0):


V(1) = (1 – 2*1)*ln((1 – 1)/1) = -1*ln(0) = infinite


But let’s say I have a half-and-half mixture of red and white marbles (x = 0.5):


V(0.5) = (1 – 2*0.5)*ln((1 – 0.5)/0.5) = 0*ln(1) = 0*0 = 0

Then my mixture doesn’t have any value at all?

Here was my first clue that there was something intuitive going on with the value function. Why does a half-and-half mixture have no value at all? Well, think about it. If you have a half-and-half mixture of marbles, you’re no closer to having a mostly white mixture than you are to having a mostly red mixture. It all depends on which kind of marble you start picking out.

But if you have a mixture that’s mostly white, or a mixture that’s mostly red, then those mixtures already have some “value”, and it’s easier to increase their “value” by removing the minority component. When it comes to enriching lithium, that’s exactly what we do. We remove more and more lithium-6 until the mixture is nearly all lithium-7.

But with uranium, surprisingly, we move the other direction on the value function. We “decrease” the value of the mixture as we “enrich” it! Hard to believe, huh?

Consider natural uranium. It’s only got 0.71% uranium-235. Let’s imagine that in white and red marbles:


V(0.0071) = (1 – 2*0.0071)*ln((1 – 0.0071)/0.0071) = 4.870

Now let’s imagine this “enriched” up to 3.5% red marbles, like enriching uranium up to 3.5%:


V(0.035) = (1 – 2*0.035)*ln((1 – 0.035)/0.035) = 3.085

Wait a minute! Are you telling me that natural uranium with a value of 4.870 is “better” than enriched uranium with a value of 3.085?

Yup, because these equations don’t say anything about whether what you’re separating is good or bad or cheap or valuable. They just describe how “separated” it is. And natural uranium, with only 0.71% U-235, is more “separated” than enriched uranium at 3.5%.

Kinda wild, isn’t it?

What about the tails? What are they worth?


V(0.003) = (1 – 2*0.003)*ln((1 – 0.003)/0.003) = 5.771

HAH! So the tails are worth the most of all, according to the value function! All of this seems kind of counter-intuitive, but if you think about it for a moment, the tails are much closer to being a pure mixture (in this case of U-238) than either the feed or the product is, so they have the highest value function.

Now of course, we would not consider 50% enriched uranium to have a value of zero (in fact we would consider it QUITE valuable) but that’s how the equations come out. And that’s because 50% enriched uranium would be neither more U-235 or more U-238, but rather a mixture that could go either way.

Isn’t it interesting that when you plug all this into the SWU equation you get something that makes sense? It still amazes me, and I wrote the dang program that shows it.

Well, that’s probably enough fun with marbles and value functions for one night–I hope you liked it! Let me know if you did…

Posted in Isotope Separation | 6 Comments »

Congratulations to Urenco USA!

Wednesday, August 18th, 2010

Boy, sometimes you think you know about something. You think you’ve got your finger on the pulse of what’s up. But you don’t.

I was very chagrined when yesterday I was relaxing in a fold-out chair during my daughter’s soccer practice, reading my latest issue of “Nuclear News”, and lo and behold I read that Urenco USA had started up their new uranium enrichment plant outside of Eunice, New Mexico!


I couldn’t believe it! I’ve been writing a multi-part series about isotope separation and uranium enrichment on this blog and I completely missed that for the first time in 50 years the United States has a new uranium enrichment plant! And to get even better, it employs a new advanced technology: the gas centrifuge, which promises vast efficiency improvements and far less power consumption than the previous way (gas diffusion) that we’ve done enrichment.

So let me say, congratulations to the Urenco USA team on their milestone and I wish them the best of success and they bring additional enrichment cascades online at their new enrichment plant!


Let me tell you another really cool thing about this new facility. About 30 miles away, west of Hobbs, New Mexico, another company called International Isotopes, Inc., is building their own fluorine recovery facility, where they will take the depleted uranium hexafluoride (DUF6) that is the main byproduct of uranium enrichment and they will recover the fluorine. They’ll recover it as anhydrous hydrofluoric acid, which is exactly the stuff that you need to convert uranium oxide ore to uranium hexafluoride in the first place. That means that between Urenco USA and International Isotopes, they’ll be following the classic dictum to “reduce, reuse, and recycle”.

They’ll reduce the accumulation of depleted uranium hexafluoride. They’ll reuse the fluorine that was used to convert uranium oxide to uranium hexafluoride in the first place. And they’ll recycle the fluorine into additional uses preparing more uranium for enrichment.

I think that’s really cool. I love it when people use uranium and fluorine together to make nuclear power. Hopefully one of these days we’ll get to do it much more directly and shave a few complicated steps off the nuclear fuel cycle!

Posted in Depleted Uranium, Isotope Separation | 4 Comments »

Enrichment, or how I learned to stop worrying and love the SWU, part 4

Sunday, August 15th, 2010

Sorry it’s been so long since I last posted, but since I got on this topic I’ve had to “run the numbers” a fair number of times, and each time that I do that, I remember that we have computers that are a lot better at that sort of thing than I am, so after a little while I break down and start writing a code to “run the numbers” for me. And codes, as you probably know, are no smarter than the guy who writes them, so it ends up taking longer than I think it will.

Nevertheless, here’s a little code I wrote to go along with this series on enrichment. It runs in Java, and most people probably already have the stuff on their computers to make it work. It really shouldn’t matter if you’re running on a PC or a Mac or a Linux or Unix box. That’s the beauty of writing things in Java. And if for some reason you don’t have Java then you can pop over to java.sun.com to get it for free.

Anyways, the code incorporates the results of the equations that I presented in the last two parts of this series along with a few more that I may or may not get into tonight. The computational part of the code took me about five minutes to write. The pretty graphics and the user interface took much much longer. So I hope you like those. They’re there to make things easier for you at the expense of making them harder for me.


What the code shows is a little picture that looks like a Pac-Man eating some cheese. In the default case (isotopic separation of uranium), the Pac-Man represents the fraction of the feed that ends up in the tails. Which is most of it, which is why it looks like a Pac-Man. The thing that looks like a little cheese wedge is the fraction of the feed that ends up in the product. That part wasn’t all that hard to get to work, but what really took a while was the clever little graphics that show a pattern of dots in the midst of blue. Those dots are in proportion to how much of the enriched isotope is in the feed or the product. You can use the slider controls on the right-hand side of the program to change the enrichment values of the tails and product and you’ll see the density of those little dots go up and down in proportion.

I hope you like that. It took a while.

Anyway, you can see just by looking at Pac-Man and his little cheese that in a typical case of uranium enrichment (from natural levels to 3.5% or so) that most of the feed ends up in the tails, and a smaller fraction ends up in the product.


Let’s consider another case of enrichment—near and dear to my heart as an advocate for thorium and fluoride reactors: lithium. In the case of lithium, the isotope we want is lithium-7, which makes up more than 90% of natural lithium. But the problem with lithium is that we REALLY need to make sure that we get that pesky lithium-6 out of our mixture. Even a little bit of lithium-6 will make real trouble for a fluoride reactor. So when you switch the material over to lithium, the product slider in on a logarithmic scale, from 99% to 99.9999%. You’ll see that effect in the “dot density” on the picture and also in the amount of SWU’s required to reach that level of purity. As far as the tails enrichment of lithium, I’m not really sure what it should be, but I know what the product should be.

Another isotope of interest for future chloride reactors is chlorine. Chlorine-37 is the less common (~24% of natural chlorine) but more useful isotope for chloride reactors, and it is likely we’ll need to use high purity chlorine-37 in those reactors. Again, I’ve used a log-scale for the product from chlorine enrichment and a normal (linear) scale for the tails enrichment. For both lithium and chlorine I threw in some guesses as to what the feed and separative work might cost, but the reality is that I have no idea what those values should be. If anyone does know I’ll be more than happy to change the code to reflect those values.

Finally I looked at hydrogen enrichment in order to recover deuterium. This a really interesting case, because deuterium is such a tiny fraction of natural hydrogen that the feed-to-product ratio is CRAZY high. You have to go through a massive amount of hydrogen to recover a little tiny bit of deuterium. You won’t even see it on the graph—it’s so tiny. So it takes a lot of SWU’s and a lot of feed to recover deuterium, and deuterium is very valuable. A liter of heavy water (D2O) costs about $5000 the last time I checked.

Anyway, have a little fun with the program and I’ll tell you more about some of the other interesting aspects of enrichment that I have learned about in upcoming posts.

Posted in Isotope Separation | 12 Comments »

Enrichment, or how I learned to stop worrying and love the SWU, part 3

Monday, August 9th, 2010

OK, so in the last post we talked about the feed-to-product ratio and how it depended only on the levels of enrichment you chose for the product and the tails. Tonight we’ll begin to get into separative work.

When I first heard of separative work, I thought it was a very abstract and obscure concept. You read about it in a nuclear engineering text and they’ll talk about how it is the reduction of entropy in a mixture or something like that, and of course it is, but I think it is much more understandable than that.

Separative work is all around us. My first experience with separative work probably came about when I was a kid and I talked my mom into buying me some Lucky Charms. You know you’ve done it—I poured the cereal and immediately began to undertake some separative work on the delicious mixture of little marshmallows and less-exciting brown cereal bites. Separative work on Lucky Charms gave you happiness in the beginning, but you also know of the depressing moment when all you have left from your separative work are the soggy “tails” of the activity.

Later on, I realized that things would end a lot better if I ate the brown cereal at the beginning, thus “enriching” my Lucky Charms in little marshmallows. Then, after an appropriate increment of restraint, I could indulge myself in the culinary joy of highly-enriched Lucky Charms, and the sugar rush would send me running down the road to school.


Last night, I observed my little daughter, who is not quite two and has not yet been formally schooled in the principles of separative work and enrichment, nonetheless showing an effective grasp of the principles when I let her have a handful of Lucky Charms. She immediately set to work—separative work—extracting the marshmallows from the handful of cereal and casting the “tails” on the floor of the kitchen. So I have high hopes for her future as a nuclear engineer based on her success with separative work thus far.

The subtitle of this series has been “how I learned to quit worrying and love the SWU”, and SWU (pronounced “swoo”) is short for “separative work unit”, and it is a way to describe in a quantitative way this process of separative work. To separate one isotope from another in a mixture, we undertake separative work. Just how much separative work is required can be calculated using nothing but values of enrichment again, which is very handy. To calculate separative work, we use a function called the “value function”, which is based only on the value of enrichment.

V(x) = (1 – 2x)*ln((1 – x)/x)

For those of you, who like me, occasionally require some mathematical refreshers, the statement “V(x)” is just mathematical shorthand for saying, “here is some value we call V, and it only depends on x, so whatever x is, just plug it in all the places you see an x on the right-hand side of the equation, and you’ll get an answer.”

So, if x was 0.03, for instance, we’d just plug in the value of 0.03 everywhere we see an x and calculate the answer:

V(0.03) = (1 – 2*0.03)*ln((1 – 0.03)/0.03) = 3.2675

The value function basically tells you the value of the mixture that you have. Surprisingly, if you have a mixture that has an enrichment of 50% (0.5), according to the value function, it has a value of zero! See, take a look:

V(0.5) = (1 – 2*0.5)*ln((1 – 0.5)/0.5) = (0)*ln(1) = 0*0 = 0

One might protest and say that a 50% enriched mixture of uranium is QUITE valuable, and in the real world one would be correct, but from the perspective of separative work, the least valuable mixture of all is the one where there is neither more or less of one component or another. And this makes sense if you stop and think about it for a moment. To have more or less of one component means that SOME separation has happened, therefore that mixture must have some greater value than a mixture of equal parts. So when we plot the curve of value function versus enrichment level, it looks like a deep valley, with the very bottom being at an enrichment level of 50%.


In order to calculate how much separative work it takes to get from one level of enrichment to another, we need to know the value of the value function for each of the three streams in the enrichment problem: feed, product, and tails. Now that we know the value function, calculating these values for each of the three streams is fairly straightforward:

V(xf) = (1 – 2xf)*ln((1 – xf)/xf)
V(xp) = (1 – 2xp)*ln((1 – xp)/xp)
V(xt) = (1 – 2xt)*ln((1 – xt)/xt)

Where xf, xp, and xt are the enrichment values of the feed, product, and tails respectively. Remember when using enrichment values in the value function, do not input them as percentages! You will get the wrong answer! You need to input them into the value function as decimal values, not percentages.

Given that you have calculated the value function for each of the three streams, you can then calculate the separative work by a knowledge of the mass in each stream:

separative work = mp*V(xp) + mt*V(xt) – mf*V(xf)

The separative work is simply the mass of the product (mp) multiplied by its value function, plus the mass of the tails (mt) multiplied by its value function, minus the mass of the feed (mf) multiplied by its value function.

But I had said earlier that I could show that separative work was only a function of enrichment levels, right? To do so just takes a few more steps.

We know that feed, product, and tails are all related.

mf = mp + mt

So it’s not hard to solve for one of these and incorporate it into the separative work calculations. Let’s solve for tails and substitute it into the separative work expression:

mt = mf – mp

separative work = mp*V(xp) + (mf – mp)*V(xt) – mf*V(xf)

separative work = mp*V(xp) + mf*V(xt) – mp*V(xt) – mf*V(xf)

Now we can collect the terms that have to do with the mass of the product and the mass of the feed:

separative work = mp*(V(xp) – V(xt)) + mf*(V(xt) – V(xf))

But recall from our last discussion that the mass of the feed and the mass of the product can be related by the feed-to-product ratio, which depends only on enrichment.

mf/mp = (xp – xt)/(xf – xt)

Let’s call this ratio “FPR” for feed-to-product ratio. mp*FPR = mf, right? So let’s substitute that into the expression for the mass of the feed:

separative work = mp*(V(xp) – V(xt)) + mp*FPR*(V(xt) – V(xf))

Now we can simply divide through by the mass of the product, and get the amount of separative work required per unit of product:

separative work/mp = (V(xp) – V(xt)) + FPR*(V(xt) – V(xf))

And that expression depends only on the enrichment levels of the feed, product, and tails.

In the next post we’ll get into some specific cases where we use these expressions in real-world scenarios.

Posted in Isotope Separation | 4 Comments »

Enrichment, or how I learned to stop worrying and love the SWU, part 2

Saturday, August 7th, 2010

Okay, before we get into the techniques of enrichment, let’s spend a little time with the fun part—the numbers! How much does it take and how much does it cost? Here’s where I’ve really enjoyed learning more lately.

The basic equations that describe enrichment aren’t that hard to derive. Even I can do it! So here goes:

You know that you’re going to start out with some amount of input material—the “feed” as it is called. You know that after you’re done you will have split the feed into two: the “product” and the waste, which in this case is called the “tails”. Assuming that you didn’t mess up and lose lots of material along the way, you could add up the product and the tails and that would be the same amount of material that you started with. So we’ll call the mass of the feed “mf”, we’ll call the mass of the product “mp”, and we’ll call the mass of the tails “mt”. And we’ll write a really simple equation that relates all three.


mf = mp + mt

Next, let’s assume that we have a mixture that has only two components. They might be U-235 and U-238, or Li-6 and Li-7, or Cl-35 and Cl-37. This derivation works for any two-component mixture. Let’s say that the fraction of the less abundant component (U-235 in the case of uranium) is given by a non-dimensional variable called “x”. And let’s say that that value of x is different for the feed, product, and tails. For instance, in the case of uranium, if the feed is natural uranium, 0.0071 of it is U-235. So the value for x for the feed would be 0.0071. xf = 0.0071. And let’s say that we want to make fuel for a light-water reactor out of this, and it needs the uranium enriched up to 3%, or 0.03. So the value of x for the “product” would be 0.03. xp = 0.03. The last thing to figure out would be how much U-235 we will tolerate in our “tails”, the part that we’re going to throw away.

At first blush, we might think, “hey, I don’t want to throw away any U-235, I want it to all go into my product. So I’ll set the tails to 0% and everything will be great.”

Hah—that seems like a good idea but it isn’t. Because the amount of effort it will take to strip every single last atom of U-235 from the mixture is infinite. So in reality don’t do that. So then you might think, “OK, that might be too hard, maybe I should set the tails enrichment higher so I don’t have to do as much work. I’ll set it to just a little bit less than the natural level of enrichment, something like 0.006%.”

Another bad idea. If you set the tails enrichment too low, you expend too much effort trying to get some small amount of stuff. If you set it too high, you end up throwing away most of the stuff you’re trying to get in the first place. So you have to be kind of careful about how you set your tails enrichment. Which will have everything to do with the economics of enrichment, as we shall see.

So I’ve spent this time telling you that you know the enrichment of the feed (easy, it’s the natural enrichment) and the enrichment of the product (easy, it’s what your customer wants) but the real question is what is the enrichment of your tails. That’s going to have to be a choice based on economics, and as economics change you might find yourself revisiting that decision.

OK, so let’s say we’ll set the tails enrichment at 0.003. xt = 0.003. That’s means about half of the U-235 in the original uranium is going to end up in your product and about half is going to end up in the tails. If we assume that the total amount of U-235 is constant, then we can write another equation:


xf*mf = xp*mp + xt*mt

This is just another way of saying that all of the U-235 ends up in either the product or the tails, and if you added it all up it would be the same as the amount that you had in the feed. Now we can combine these two equations to start figuring things out. We can use some of those tricks we learned in eighth-grade algebra to solve for things when we have two equations. We can rewrite the first equation to equal the tails rather than the feed:


mt = mf – mp

And we can substitute that definition into the second equation:


xf*mf = xp*mp + xt*(mf – mp)

Then, again using our eighth-grade algebra skills we can expand the equation:


xf*mf = xp*mp + xt*mf – xt*mp

and we can group the terms relating to the mass of the feed and the product:


xf*mf – xt*mf = xp*mp – xt*mp

We pull out the common factors…


mf*(xf – xt) = mp*(xp – xt)

…and then we can solve for the mass of the feed:


mf = mp*(xp – xt)/(xf – xt)

Or even better, we can figure out a ratio between the mass of the feed and the mass of the product:


mf/mp = (xp – xt)/(xf – xt)

Hooray! So why is this a big deal? Because now we have an answer that doesn’t really care how much actual mass of feed or product that we’re talking about. On the left-hand side of the equation is a ratio, and on the right-hand side of the equation are a bunch of enrichment values. Let me show you how useful this nifty little expression is.

Let’s say that you want to figure out how much natural uranium you will need to fuel a nuclear reactor that uses enriched uranium. This is pretty much the situation we are in in the United States. Let’s say that you know that a typical reactor takes about 35 tonnes of enriched uranium fuel per year, and that that fuel is enriched up to 3%. You already know that the enrichment level of natural uranium is 0.0071, and let’s say that you’re comfortable with your tails enrichment being 0.003. So then you run the numbers:


mf/mp = (xp – xt)/(xf – xt) = (0.03 – 0.003)/(0.0071 – 0.003) = 0.027/0.0041 = 6.58

See how the tails enrichment shows up in both the numerator and the denominator of the expression? In this case, the ratio is 6.58, which tells you that you’ll need almost seven times more uranium as a feed material than you’ll need for fuel. So if you need 35 tonnes, you multiply 35*6.58 to get 230 tonnes. You’ll need 230 tonnes of natural uranium to make 35 tonnes of enriched uranium. Where does the rest of the uranium go? Into the tails. There’s 195 tonnes of uranium tails in this example. One of the things that you find when you do uranium calculations is that you almost always make a lot more tails than you make product.

But let’s do another example, to broaden our perspective. Let’s say that we’re talking about enriching lithium now instead of uranium. We want to make lithium-7 for a LFTR, and we need it to be enriched in lithium-7 up to a point of about 0.9999 (the more nines the better). But natural lithium is only about 90% enriched in lithium-7. So how much lithium feed will we need to enrich to 0.9999? Again, it has everything to do with the enrichment we’ll tolerate in the tails. Let’s say (being foolish) that we set the tails enrichment to 0.80. Don’t want to work too hard, right? So let’s go run the numbers and find a feed-to-product ratio for our lithium enrichment.


mf/mp = (xp – xt)/(xf – xt) = (0.9999 – 0.8)/(0.9 – 0.8) = 0.1999/0.1 = 1.999

In this case, we need about twice as much lithium feed to get a particular level of product. Why is that? Because by setting the tails to 0.8 only about half of the lithium-7 ended up in the product. The other half if in the tails. So even in a case where we have something like lithium-7, where it is the dominant constituent of natural lithium, setting the tails level of enrichment has a lot to do with how much feed it will take.

So let’s go back and run it again, this time with a tails enrichment of 0.5. No sense letting all that valuable lithium-7 go to waste, right?


mf/mp = (xp – xt)/(xf – xt) = (0.9999 – 0.5)/(0.9 – 0.5) = 0.1999/0.1 = 1.25

Now we only a little bit more lithium feed (25%) than we expect as product—in other words, most of the lithium-7 ended up in our product stream rather than in the tails. Seems like a better way to go, right? Well, maybe. It still depends on a lot of other things, like how hard it actually is to do the separation.

And getting into that level of detail will be a subject of an upcoming post…

Posted in Depleted Uranium, Isotope Separation, Lithium | 1 Comment »

Enrichment, or how I learned to stop worrying and love the SWU, part 1

Friday, August 6th, 2010

“Enrichment” or more particularly “uranium enrichment” is probably one of those phrases that the average person hears on television or reads on the Internet and has only the vaguest concept of what it means. They likely think “it’s bad” and “it has something to do with uranium, which I think is bad, for some reason”. Commentators and journalists decry the idea that Iran *gasp* might have access to uranium enrichment. Our leaders both domestic and international, loudly thunder that lesser nations shall not be allowed to develop enrichment.

When Paul got to Rome in Acts 28:22, the local Jews came up and asked him about Christianity, saying that they didn’t know what it was but that everything that they heard about it was bad. Likewise, the average person could ask today, “so what’s up with enrichment? All I hear about it is that it’s bad!”

The world of the nucleus is a secret world, shielded from everything else by the cloud of electrons surrounding it. The electrons do all of the interacting, the nucleons do nearly nothing. The only hint they give to their existence is their charge, which is balanced by the electrons, and their mass. Practically all of the mass of the atom is in the nucleus.

Different isotopes of elements have different masses. They have the same number of protons, and hence the same charge and the same chemical nature, but different masses. From a chemical perspective, there’s no difference between most isotopes (with a few exceptions, like normal hydrogen and deuterium). But from a nuclear perspective there can be a huge difference between two naturally occurring isotopes. Lithium is a good example. Natural lithium is 90% lithium-7 and about 10% lithium-6. Chemically they’re the same. You can make batteries from lithium-6 just as well as you can make batteries from lithium-7. But from a nuclear perspective, they’re really, really different. Lithium-7 has almost no appetite to absorb neutrons, whereas lithium-6 has a HUGE appetite to absorb neutrons.

Uranium is in an analogous situation. Uranium-235 and uranium-238 are the two naturally occurring isotopes of uranium. One of them is far more common than the other. Can you guess which? Yeah, it’s the one that we’re not so interested in. Kind of like how when your toast falls down it always seems to land on the buttered side. Nevertheless, we’re a lot more interested in uranium-235, which is only 71 out of every 10000 uranium atoms than we are about uranium-238, which is the other 9920 out of 10000 uranium atoms. (1 atom out of 10000 is uranium-234, but we won’t worry about that right now).


Enrichment is the process where you separate one kind of isotope from another, and the overwhelming reason to do it is because you intend to use your product in a nuclear process. Because if you want to do a chemical process, you usually don’t care too much about what kind of isotope mix you have. Sometimes, but very rarely.

So how do you conduct enrichment? Well, your toolbox is pretty sparse. All of the chemical tricks that are so commonly used to separate one thing from another don’t work when you’re dealing with one kind of element, because chemically it’s all the same. You have to try to separate one thing from another based on something that’s different between the two.

Almost always that difference is mass. Lithium-6 has six units of atomic mass, lithium-7 has seven. Uranium-235 has 235 units of atomic mass, uranium-238 has 238. So there’s a difference there. It’s more pronounced for lithium than uranium, but it’s there nevertheless.

Separating uranium is a big deal because U-235 is fissile and U-238 is not. The original motivation to separate uranium was to make a bomb during the Manhattan Project. The first nuclear bomb ever used in war, 65 years ago today, was based on separated uranium-235. And enrichment has had a bad rap ever since.

To separate U-235 from U-238, you need to put them in a chemical form where it’s easier for them to get away from one another. Thank goodness for fluorine! Uranium hexafluoride (UF6) is a gaseous form of uranium. It has one uranium atom in the middle of six fluorine atoms. It looks a lot like one of those little jacks my sister used to play with. There’s another aspect to UF6 that’s really important. If fluorine itself had a number of different isotopes then this whole idea wouldn’t work. For instance, if there were isotopes of fluorine that had mass numbers of 18, 19, and 20, then you could go build a UF6 molecule and one of its fluorine’s might be 18, and another would be 20 and another could by 19, and that mass difference from uranium could get “lost in the noise” from all of the different isotopes of fluorine.

But fortunately, that’s not how it works. Fluorine has only one natural isotope, number 19. So when you go add six fluorines to uranium, you know that they’re all going to be 19. Any difference in mass from the molecule all comes from the uranium, not the fluorine. So we’ll do a little math. If you have a uranium atom that has a mass of 235, and you go and add six other atoms each with a mass of 19 to it, what do you get? (235 + 6*19) = 349. (I won’t lie, I used my calculator) And if you have a uranium atom with a mass of 238? (238 + 6*19) = 352.

So the whole trick in uranium enrichment is to separate molecules in a gas that have a mass of 349 from molecules that have a mass of 352. Folks, that’s not very much, and you can see why uranium enrichment is hard and expensive. But, thanks to the gaseous nature of uranium hexafluoride and the monoisotopic existence of fluorine, it’s possible in the first place, where otherwise it might not be.

Next post, how to separate 349 from 352…

Posted in Depleted Uranium, Isotope Separation | 13 Comments »