In this section, you will learn the definition of these terms
- two-stage decay chain
and you will use information from previous sections about
- decay constant
In a previous section we talked about simple radioactive decay, where a radionuclide decayed into a stable nuclide. While this is common, it is also common that there are actually two steps in a radioactive decay sequence. First the radionuclide decays into another radionuclide, and then decays into a stable nuclide.
An example of this type of decay is found in one of the products of nuclear fission, iodine-133. Iodine-133 has a half-life of 20.8 hours and decays into xenon-133, which is also radioactive and has a half-life of 125.8 hours. Xenon-133 decays into stable cesium-133, which is the only stable isotope of cesium found in nature.
Let’s suppose that some number of iodine-133 atoms are present initially. We’ll call this amount N01 with the 0 subscript meaning “initial amount” and the 1 subscript meaning “the first stage of the decay chain”. We can write a differential equation (an equation describing how something changes) to represent the change in the number of iodine-133 atoms.
This differential equation may look familiar from the lesson on simple radioactive decay. It is the same equation we used before to describe simple radioactive decay with one small modification. We’ve added the “1″ subscript to distinguish the number and the decay constant as belonging the first stage of the decay chain.
Now let’s try to write a differential equation for the second stage of the decay chain—the formation and destruction of xenon-133.
Unlike the differential equation for the iodine-133, this differential equation has two terms on the right-hand-side of the equation. The first one is a “formation” term, and it represents the xenon-133 being generated as iodine-133 decays into it. That’s why it has the “1″ subscript, because it concerns the first stage of decay. It also has a positive sign because as the first stage decays it forms the second stage. The second term is the “destruction” term and it concerns the xenon-133 as it decays away. That’s why it has the “2″ subscript and has a negative sign, because as xenon-133 decays it is turning into something else.
The third stage is pretty simple. Cesium-133, since it is stable, only has a formation term and no destruction term. Xenon-133 (designated by the subscript 2) is decaying and forming cesium-133, which is stable and won’t decay away.
We can solve these differential equations and the first stage simply becomes the expression for simple radioactive decay, with the “1″ subscript designating that it is the first stage of the decay chain.
Just like in simple radioactive decay, if we know the initial number of the nuclei in the first stage, we can solve for the number at a later time, given the time interval and the decay constant. Remember we can easily recast this equation in terms of activity instead of number without difficulty, with A01 becoming initial activity instead of initial number.
Solving for the second stage of decay is more complicated, and the derivation of this equation is not given here but the result is given, both for number and activity.
The amount of the stable third stage of the decay sequence is obtained directly from a material balance. Basically we assume that since the nuclei don’t go away as they decay, if they aren’t at stage 1 or stage 2, they must be at stage 3.
Why can’t we calculate activity for the third stage? Because the nuclide is stable and it has no activity. Its decay constant is zero. So activity has no meaning for a stable nuclide and we can only use number.
Example 1. Barium-140, a fission product (with a half-life 12.8 days), decays to lanthanum-140 (with a half-life of 40.5 hours), and then lanthanum-140 decays to stable cerium-140. Starting with 1.0 milligram of freshly purified barium-140 plot (a) the amount of each isotope and (b) the total activity in curies, as functions of time. (Be sure to extend the plots beyond any maxima in the curves.)
Barium-140 has a decay constant of ln 2/((12.8 days)*(86400 seconds/day)) = 6.27 x 10-7/sec. With an atomic mass of 140, it has a specific activity of (6.27 x 10-7/sec)*(0.6022 x 1024)/(140 g) = 2.70 x 1015 Bq/gm or 72.9 kCi/gm.
Thus one milligram of barium-140 has an activity of 72.9 curies. Lanthanum-140 has a decay constant of ln 2/((40.5 hours)*(3600 seconds/hour)) = 4.75 x 10-6/sec.
Knowing the initial activity of the one-milligram barium-140 sample (72.8 curies) and the two decay constants allows us to use a spreadsheet program to compute the activities of barium-140 and lanthanum-140 over a period of time using the expressions for first stage and second-stage radioactive decay. Here’s a plot of the results.
As you can see from the graph, the barium activity starts out at around 73 curies and decays from there. The lanthanum activity, initially zero, rises and rises until it reaches an activity equal to that of its parent (and there’s an interesting reason for that) and then begins to fall as well. Isn’t it interesting that the lanthanum activity is higher than the barium activity?
If you’d like to look over my spreadsheet, here’s a link to it.