*Dear Dr. Zoomie – I am trying to brush up on some of my radiation knowledge and am having some trouble figuring out some of the calculations and concepts about radioactivity. For example, I know there’s some sort of relationship between a nuclide’s half-life, its mass, and the amount of activity per gram but I’m not quite sure how these all go together. Can you help explain how it all works?*

You’re right – these factors are all tied together, and the relationship is fairly straightforward. That’s the good news – but it will take going through a bit of math to see how these all fit in with one another. Luckily it’s not too involved, so let’s walk through it a step at a time. First, let’s start with some equations!

*Decay constant (λ)*

The first concept is something called a decay constant, which is represented mathematically with the Greek letter lambda (λ). The decay constant is simply the probability that given atom will have a radioactive decay in a particular amount of time. Or, if you have a bunch of radioactive atoms, the decay constant tells us how many of those (what percentage of them) will decay away in a given amount of time. The equation is . The natural logarithm of 2 (ln 2) is roughly equal to 0.693 and t_{1/2} is the half-life of the nuclide you’re calculating activity for. So for Co-60 (which has a half-life of 5.27 years) the decay constant is equal to 0.693/5.27 years = 0.1315 yr^{-1}, which you would read as 0.1315 per year. What this tells us is that any particular Co-60 atom has a 13.15% chance of decaying during the course of a year, or that 13.15% of the atoms in a bunch of Co-60 will decay during a year.

The next part of this is to turn this into a measure of radioactivity. Radioactivity measures the number of radioactive decays an amount of radioactivity will undergo in a second. One curie of radioactive material will undergo 37 billion decays every second; one Becquerel of radioactivity will undergo 1 decay every second. The way to find out how many atoms will decay in a given amount of time is to count (or calculate) the number of radioactive atoms and to multiply this by the fraction of those atoms that will decay in a second. Mathematically, this is written as A= λN where A is the amount of radioactivity (decays per second) and N is the number of radioactive atoms.

OK – so say you have a billion atoms of Co-60 and you want to figure out how much radioactivity this represents. Using this equation, over the course of a year 13.15% of these atoms will decay, so one billion Co-60 atoms will undergo 131.5 million radioactive decays, giving a decay rate of 131.5 million decays per year. This is easier to calculate with if we use scientific notation – written this way we’ve got 1.315×10^{8} decays per year. Unfortunately, our instruments don’t read out in counts per year so we have to convert this to a more useful measure. There are roughly 31.4 million (3.14×10^{7} seconds in a year, so we divide 1.315×10^{8} decays per year by 3.14×10^{7} seconds per year to get a decay rate of about 4.2 decays per second (dps), which is about 4.2 Bq, measuring radioactivity in SI units. To convert to the US units of curies we have to remember that one Ci is 37 billion Bq, so 4.2 Bq = 1.13×10^{-7} mCi, or about 113 pCi (a pCi, or pico-curie, is a trillionth of a curie. So a billion atoms of Co-60 gives us 113 pCi of radioactivity.

Now, let’s see what happens when you have a nuclide with a different half-life. What if, instead of Co-60, you have a billion atoms of Cs-137 (which has a half-life of about 30 years)?

First, the decay constant for Cs-137 is going to be different; so the same billion atoms will produce only 23.1 million decays per year, compared to the 131.5 million Co-60 decays. So Cs-137, which has a longer half-life, decays more slowly than the shorter-lived Co-60. And this is a good rule of thumb – for the same number of radioactive atoms, the ones with the longer half-lives will decay more slowly (will have lower levels of radioactivity). And as a corollary, nuclides with shorter half-lives will be more intensely radioactive for the same number of radioactive atoms.

*Specific activity*

OK – so this tells us how half-life and radioactivity go together. But we don’t count atoms (at least, not normally) – weight is how we normally measure things. So a more useful measure – a more useful calculation to perform – will tell us how much radioactivity we have per gram of material; the name for this is “specific activity.” And when we look at specific activity we have to take into account not only a nuclide’s half-life, but also how massive the atoms are. This gets a little more complicated, but there’s a bit of a shortcut at the end that makes things a bit easier.

Say we have one gram of Co-60. If we want to use the equation we already know then we have to figure out how many atoms of Co-60 there are in a gram. This is the somewhat complex part. Going back to high school chemistry, remember that the mass of one mole of anything (the number of grams that it weighs) is equal to the molecular (or atomic) mass. The mass part is easy – it’s just the numerical part of the nuclide. So one mole of Co-60 weighs 60 grams (just as one mole of Cs-137 will weigh 137 grams, and one mole of Ra-226 will weigh 226 grams). And one mole of Co-60 (or anything else, for that matter) has 6.022×10^{23} atoms, a number called Avogadro’s Number, after the 19^{th} century scientist who first calculated it). So the number of atoms in one gram of Co-60 is equal to one sixtieth of Avogadro’s number, or atoms of Co-60. And from here, we just multiply by the decay constant, which we know from earlier to be 0.1315 per year. Doing this tells us that one gram of Co-60 will undergo about 1.315×10^{21} decays per year, or about 4.2×10^{13} decays every second. Remember that one curie will undergo 37 billion (3.7×10^{10}) decays per second: So one gram of Co-60 gives us 1132 Ci – this is the specific activity of Co-60.

Using the same reasoning as before, we can guess that a nuclide with a longer half-life should have a lower specific activity since fewer atoms are decaying every second. And adding to that, a heavier radionuclide should also have a lower specific activity (fewer Ci or Bq per gram) because there are fewer atoms in a gram when each atom is heavier.

So here’s how it all goes together. All else being equal:

- A shorter half-life means a more radioactive nuclide (more Ci or Bq for the same number of atoms)
- A longer half-life means lower activity (fewer Ci or Bq per gram for the same number of atoms)
- A heavier nuclide (the number part of the nuclide is larger) means fewer atoms in a gram, so there’s less radioactivity per gram
- A lighter nuclide means more atoms per gram, so a higher specific activity
- And if both half-life AND weight change then you have to figure out which is more significant before you can tell.

*An easier way to do the math*

OK – that’s the “pure” way to figure this out, but as I promised earlier, there’s an easier way to do it – just compare to a radionuclide with a known specific activity. And the one I compare against is the nuclide that gave us the definition of 1 Ci – Ra-226.

One gram of Ra-226 has an activity of 1 Ci (not exactly, but close enough for our purposes) and Ra-226 has a half-life of 1600 years. So a radionuclide with a similar mass and a shorter half-life will have a higher specific activity. Consider Am-241, with a half-life of about 432 years and a mass that’s fairly close to that of Ra-226. Roughly speaking, Am-241 has a half-life that’s about a quarter as long as Ra-226 so we’d expect to see four times as many atoms decay in the same amount of time. This means that 1 gram of Am-241 should have a little less than 4 Ci of activity. And if we look it up, we find that Am-241 has a specific activity of 3.4 Ci/gm – right in the ballpark of what we predicted.

Another one we can try is Co-60. In this case we have to correct for both the mass and the half-life:

Ci/gram

Remember – a longer half-life means a lower decay constant (fewer atoms decay in the same amount of time) and a heavier atom means that there are fewer atoms in a gram of the pure nuclide. This is very close to what we calculated earlier, close enough to not worry about the relatively minor discrepancy that comes from the rounding off we’ve been doing.

*Wrapping up*

After having gone through all of that I have to admit that there aren’t times you’ll need to go through this, and there aren’t many radiation safety officers who need to make these calculations. Those who manufacture radioactive sources need to know this of course. If you’re working in a nuclear pharmacy or in radiation oncology then these calculations will also be useful. And if you’re a scientist using radionuclides for your research then these are calculations you need to be able to perform. And, I guess, if you hang out at geek bars, this particular skill set might win you a free drink (or, then again, maybe not…). But if you’re not in one of these groups (and if you’re not studying up for an exam of some sort) then you can probably get by with using a spreadsheet for all of this.