# Help with population of various isotopes in decay chains (Bateman equations)

1. Nov 30, 2010

### alby

(Apologies for cross-posting this in the Nuclear and Differential Equations forums. I'm new to posting on Physics Forums and not entirely sure where it belongs. Mods, please feel free to delete/move as appropriate.)

I am trying to create a decay chain simulator in Excel that my pupils can use to create graphs similar to those created by the Nucleonica Decay Engine: the idea being that they can find a decay chain and enter the isotopes and their half-lives and Excel will create the graph.

The problem I'm having is with the equations required to calculate the population of each isotope at each stage in time. My initial approach was far too simplistic and whilst it created correct *looking* graphs, the numbers weren't correct.

As an example, I'm using the last four stages of the actinium series uranium decay chain:
$${}^{211}\mathrm{Bi}\rightarrow {}^{211}\mathrm{Po}\rightarrow {}^{207}\mathrm{Tl}\rightarrow {}^{207}\mathrm{Pb}$$

I can calculate the population of the Bi-211 at time t with:
$$N_{Bi}=N_0 e^{-\lambda t}$$

and the population of Po-211 using:
$$N_{Po} = \frac{\lambda_{Po}}{\lambda_{Bi} - \lambda_{Po}} N_0(e^{-\lambda_{Bi}t}-e^{\lambda_{Po}t})$$

The problem is with the equations for the population of Tl-207 and Pb-207. I know I should be looking at Bateman's work on the subject, but as a lowly teacher I can't access the literature.

2. Dec 11, 2010

### gato_

Seems like the rate equations are simply
$$\dot{Po}=-\lambda_{Po}Po+\lambda_{Bi}Bi$$
and so on. For example, the one for Ti would be
$$\dot{Ti}=-\lambda_{Ti}Ti+\lambda_{Po}Po$$
By the way, the solution I get for Po is
$$Po=N_{0}\frac{\lambda_{Bi}}{\lambda_{Po}-\lambda_{Bi}}[e^{-\lambda_{Bi}t}-e^{-\lambda_{Po}t}]$$
All of the equations are valid up to [Po], [Bi],... equal to 0. There is no negative number of samples

3. Dec 12, 2010

Sorry, but I couldn't get the tex in my reply to display properly without including these three expressions. Please ignore them:

$$1$$
$$1$$
$$1$$

I have taken the following info from [Cetnar, Annals of Nuclear Energy 33, 640 (2006)].

An expression for the population of the nth nuclide in a serial decay chain (derived by Bateman):

$$N_n(t) = \frac{N_1(0)}{\lambda_n} \sum^{n}_{i=1} \lambda_i \alpha_i \exp{[-\lambda_i t]},$$

where

$$\alpha_i = \prod^{n}_{j=1, j \neq i} \frac{\lambda_j}{(\lambda_j - \lambda_i)}.$$

Hopefully the notation is obvious -- I'm struggling to get my tex expressions to display properly in this post. The N_1 is the parent population, N_2 the daughter population, and N_3 the grand-daughter population (in your case Bi211, Po211, and Pb207, respectively).

Note that this gives a different expression for the Po population you were attempting to calculate:

$$N_{Po} = \frac{\lambda_{Po}}{\lambda_{Bi} - \lambda_{Po}}N_0(e^{-\lambda_{Bi}t}-e^{\lambda_{Po}t})$$

which agrees with that stated by Gato.

I get the impression that you are treating Tl207 as a daughter nuclide of Po211, whereas it is actually a daughter nuclide of Bi211, and so its population can be calculated in more or less the same way as the population of Po211 (I guess that to get the correct populations you have to scale the lifetime of Bi211 by the fraction of mutations that end in Po211 or Tl207; i.e., that in your calculation of N_Po you don't use $$\lambda_{Bi}$$, but instead use $$0.00276 \lambda_{Bi}$$ as only 0.276% of Bi211 mutations end in Po211 (according to a figure in http://en.wikipedia.org/wiki/Decay_chain)).

To calculate the population of Pb207 I would use the expression separately for each path down the decay chain from Bi211 to Pb207 (i.e., Bi211 -> Po211 -> Pb207, and Bi211 -> Tl207 -> Pb207), and then take the sum of the two expected populations to get the actual population of Pb207. This is just a suggestion of how it can be done; I haven't tried it, and can't say with certainty that it is correct. Do let us know if this works for you.

4. Nov 27, 2011

### HotCells

You mention the Nucleonica decay engine - but why don't just use this?. The people at Nucleonica (www.nucleonica.com) have solved all the problems you mention.

5. Nov 27, 2011

### alby

"Why bother learning to play the piano? Other people can play the piano for you."