A Few Good Moderators: The Numbers

Yesterday I posted on the issues that surround finding a good moderator. Today I’ll walk you through the math that will let you do the calculations yourself. At the end I’ll link a spreadsheet that already has the calculations done that you can download and monkey around with if you want.

First of all, you want a moderator that’s good at moderating. That means that when a neutron hits it, the neutron loses a lot of energy, specifically kinetic energy. Just like in marbles or pool balls, the bigger the moderator nucleus, the less energy the neutron will lose when it hits it.

In nuclear engineering, the fancy term for this feature is the average logarithmic energy decrement per collision. And if that’s not fancy enough, they use a funny Greek letter that looks like a squiggle to represent it. I think the letter is actually called “xi”, but I prefer to call it “squiggle” since that’s what it looks like.

Calculating squiggle for each nucleus is really easy. All you have to know is “A”, the atomic mass of the nucleus. For hydrogen-1, A is 1, for hydrogen-2 (deuterium) A is 2, for helium-4 it’s 4. You can see this is very difficult. For carbon-12, you’ll never guess what A is. Okay, stop guessing, it’s 12.

After hours of deductive logic to compute A for a nucleus, you’re ready to calculate “alpha”. Alpha is simply:

alpha = frac{(A – 1)^2}{(A + 1)^2}

So for hydrogen-1 (protium), it’s

alpha(text{hydrogen-1}) = frac{(1 – 1)^2}{(1 + 1)^2} = frac{0}{4} = 0

For helium-4 it’s

alpha(text{helium-4}) = frac{(4 – 1)^2}{(4 + 1)^2} = frac{9}{25} = 0.36

For carbon-12 it’s

alpha(text{carbon-12}) = frac{(12 – 1)^2}{(12 + 1)^2} = frac{121}{169} approx 0.716

You can see the trend–for heavier and heavier nuclei, alpha gets closer and closer to one. For hydrogen it’s zero, and everything else is between zero and one. We’ll find that moderation goes better when alpha’s more like zero than like one.

Armed with our calculation of alpha, we can calculate the logarithmic energy decremental operation of collisionary …. um, squiggle. Yeah, with alpha we can find squiggle.

And here’s the equation for squiggle:

xi = 1 + frac{alphaln(alpha)}{1 – alpha}

Let’s use a spreadsheet to spare us some of the complicated math and look at the results for a bunch of different nuclei. You can do this at home and catch all my math errors:

A alpha squiggle
hydrogen 1.000004 0.0000 1.0000
deuterium 2 0.1111 0.7253
helium-4 4 0.3600 0.4253
lithium-7 7 0.5625 0.2602
beryllium-9 9 0.6400 0.2066
carbon-12 12 0.7160 0.1578
oxygen-16 16 0.7785 0.1199
fluorine-19 19 0.8100 0.1017
sodium-23 23 0.8403 0.0845
chlorine-37 37 0.8975 0.0531
uranium-238 238 0.9833 0.0084

Thanks to the spreadsheet and it’s patience in doing tedious calculations like alpha and squiggle, I was able to quickly find out these values after only going to the trouble of typing in the name and deducing the atomic mass. Now some of you may be wondering why I typed in 1.000004 for the mass of hydrogen rather than just one, and I won’t deprive you of the fun you’ll have later on if you try to build your own squiggle-calculating spreadsheet and DON’T do this, so I won’t tell you that all your equations that use hydrogen won’t work if alpha is EXACTLY zero, so go ahead and use one for hydrogen’s mass and see what happens.

What can you do with a calculation of squiggle? Well, a few different things. One of the things you can do is that you can figure out how many collisions it takes (on average) to slow a neutron down from the high-energies at which it is born to the slow energies that improve the probability of fission.

For instance, if we wanted to find out how many collisions it took between a neutron and a moderator nucleus to slow the neutron down from 2 million electron-volts to 1 electron volt, then the equation would simply be the natural log of the ratio of the energies divided by squiggle.

Like this:

text{# of collisions} = dfrac{lnleft(dfrac{2000000 eV}{1 eV}right)}{xi} = 14.51

# of collisions = ln (2,000,000 eV/1 eV)/squiggle

ln (2,000,000/1) = 14.51

So for hydrogen, it would take 14.51/1 = 14.5 collisions, on average, to slow a neutron down from 2 MeV to 1 eV. If your squiggle’s less than one, it will obviously take more collisions. Here’s a quick list:

A alpha squiggle 2 MeV > 1 eV
hydrogen 1.000004 0.000 1.000 15
deuterium 2 0.111 0.725 20
helium-4 4 0.360 0.425 34
lithium-7 7 0.563 0.260 56
beryllium 9 0.640 0.207 70
carbon 12 0.716 0.158 92
oxygen 16 0.779 0.120 121
fluorine 19 0.810 0.102 143
sodium 23 0.840 0.084 172
uranium-238 238 0.983 0.008 1732

You can see that it takes a lot of collisions to slow down neutrons when you’ve got a lot of heavy stuff in the core, and a lot less when you’ve got light stuff like hydrogen or deuterium.

Now we still don’t know if a neutron can survive 20 or 30 collisions with a moderator and not get gobbled up! We also don’t know how to look at combinations of nuclei, such as compounds like water or beryllium fluoride or things like that.

2/14 continuation…

OK, so now we want to take on moderating compounds. I mean, knowing these element numbers is nice and all, but let’s say we want to compare heavy water to beryllium hydride…how do we know which one is better/worse/whatever?

To take the next step, we’re going to need some of the “microscopic cross-sections” for each of the moderator nuclei. Conveniently, we can find these cross-sections in a variety of different places. I happened to get these from the “Map of the Nuclides” on the LANL Nuclear Information Service site. These cross-sections are the room-temperature (2200 m/s) cross-sections from the JENDL 3-2 library.

sigmaS sigmaA
hydrogen 20.474 0.33200
deuterium 3.389 0.00055
lithium-7 0.970 0.04500
beryllium 6.151 0.00920
carbon-12 4.746 0.00350
oxygen-16 3.780 0.00019
fluorine-19 3.643 0.00960
sodium-23 3.024 0.53100
chlorine-37 1.150 0.43300
uranium-238 9.360 2.71700

I include lousy moderators like chlorine and uranium in here for comparison purposes. Each of the cross-sections is presented in the unit of “barns”. A barn is a useful little definition–one barn is one-septillionth of a square centimeter, or 10^-24 cm^2.

We use “barns” a lot in nuclear engineering–I think the name came from something like “hitting the side of a barn”. At any rate, we can look at these barn numbers and see a few different things. There are two cross-sections shown here. One is the scattering cross-section. That’s the “size” of nucleus in the sense of colliding with a neutron and “scattering” it, in other words, slowing it down and not absorbing it. This scattering cross-section is called “sigma sub s” for the way you see it written in the books: a Greek letter sigma (lower-case) with a little “s” as a subscript, indicating that it’s a scattering cross section.

The other cross-section is the “absorption cross-section”, also called “sigma sub a”. This is written as a lower-case Greek letter sigma with a little “a” as a subscript indicating that it’s for absorption, not scattering.

As you might have guessed, for a moderator, a big scattering cross section is GOOD, and a big absorption cross section is BAD. You don’t want moderators to absorb neutrons, you want absorbers to slow them down! So we want material that presents a big target for slowing down, and presents a small target for absorption.

But what about our old friend squiggle? Well, squiggle is how much energy is lost in one of those collisions in the first place. So we really need to think about how big of a scattering target the nucleus is (sigma sub s), how much energy it loses in the collision (squiggle), and how improbable an absorption might be (sigma sub a).

But what is squiggle for a compound? What is it for something like beryllium fluoride, where you’ve got both beryllium and fluorine nuclei in the compound? How do you find squiggle?

Well, in the case of compounds, squiggle becomes a weighted average, and the weighting factor is the microscopic scattering cross-section (sigma sub s). For instance, for something like heavy water (D2O), where you have two deuterium nuclei and one oxygen nuclei, to calculate squiggle for this compound you would multiply the squiggle for deuterium by the sigma sub s for deuterium, times two (because you have two deuterium nuclei in the compound) and add to that the squiggle for oxygen multiplied by the sigma sub s for oxygen. Times one because there’s only one oxygen in heavy water. Then divide the whole bunch by two times the scattering cross section of deuterium added to the scattering cross section of oxygen.


Here’s how it would look:

Continue the discussion in the forum

7 thoughts on “A Few Good Moderators: The Numbers

  1. Humm, you should consider though that the scattering cross section is not independent of energy. So this weighting would change from MeVs to thermal if one is exact about it.

  2. Humm, you should consider though that the scattering cross section is not independent of energy. So this weighting would change from MeVs to thermal if one is exact about it.

    Also, I don't see the spreadsheet anywhere

  3. Howdy,

    Thank you so much, the Nuclear Concepts for Engineers by Mayo did not explain the ? for compounds at all. Your explanation is amazing, and has granted me deeper understanding of, "squiggle e," or Xi.

  4. How does atomic density factor (slightly different from mass density as dense elements have less atoms per cubic centimetre for the same density) into how well a?moderator works? Obviously higher density is better solids better than liquid better than gases. Is there some limiting effects where the efficiency improvement falls with higher density?

  5. Michael – Actually, the atomic density does not change the NUMBER of collisions required for the neutron to become thermalized; instead it affects the average distance traveled by the neutron during thermalization, also known as the "slowing down length" or "mean free path" depending on if you are looking at a purely linear distance (the former) or a distance from birth to thermalization in a 3D realm (the latter). This term is useful in determining "buckling" of the core.

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