# How should I interpret these decay chain numbers?

I've been parsing JSON radioactive decay data from the IAEA, and am trying to make sure I understand the numbers.

Here is a simple example:

U-238 Decay α 100% SF 5.5E-5% Major Radiations Type keV % α 4198 79 4151 21 γ 13.00 7.3 49.55 0.064

OK. If I have this data, and I want to know how to guess what kind of radiation is going to be released over time from a given amount of U-238, what can I conclude? My understanding: nearly 100% of the time it is going to have alpha decay, and I can probably (for my purposes) just ignore the spontaneous fission. Great. 79% of the time, that alpha will be in the form of a 4.198 MeV particle, and 21% of the time it will be 4.151 MeV.

OK, but what about the gammas? Will there be a 13 keV gamma only 7.3% of the time? Is that 7.3% of the time for every alpha decay or is it only related to one of the specific alphas? This is the sort of thing I'm confused about.

One more example that illustrates some of my confusion:

Co-60 Decay β- 100% Major radiations: Type keV % β- 95.77 99.88 625.87 0.12 γ 1332.492 99.9826 1173.228 99.85

OK, so 100% of the time, this decays via beta-minus. 99.88% of the time, that beta particle is going to be 95.77 keV, 0.12% of the time it is going to be 625.87 keV. The next bit is less clear to me: does this mean that 99.9% of the time, it will have a 1.3 MeV gamma, and that 99.85% of the time it will have a 1.2 MeV gamma? That is, if I were calculating the gamma output, for each decay I'd need to check both of these gammas?

I know there are lots of gammas I am not counting here, and I'm really fine with that (for my purposes, a rough count of the most prominent ones is fine), but I just want to check that I'm understanding this correctly.

Thank you.

• It is difficult (well nigh impossible, really) to completely understand the decay dynamics just from these tables. For one thing there is no explicit correlation between various lines. In the case of Po-210 you can say "The energy of the occasional gamma is the difference of the two alpha lines, and the branching fraction for the low energy alpha is the same as that of the gamma, so they must go together." In other cases, no: the table give little hint that the two high energy gammas of Co-60 are usually emitted in coincidence (it can be inferred from a calculation involving the daughter). – dmckee Oct 1 '16 at 0:51
• I second what dmckee said. You probably want to look at the nuclear level diagrams as well - they will give you what is decaying to where, and has much of the detail you want. – Jon Custer Oct 1 '16 at 4:00
• Thanks. I am trying to process this data computationally — hence the need to just get it in some kind of numerical form. This has proven frustratingly difficult to get in bulk. – nucleon Oct 1 '16 at 12:19
• You say you want the data in a machine readable format, but perhaps you should say what you hope to do with it. This problem has been faced by Monte Carlo authors before, and while I don't know for sure what they did about it I know you can probably find out (at worst by browsing the Geant4 source). But to aid solving your problem it would help to know what your problem is (as opposed to how you thought you were going to solve it). – dmckee Oct 1 '16 at 15:35
• Come to think of it, I think I used some ENDFs for the import source of the very limited little Monte Carlo I wrote. Of course, it was so small that I performed the input by vgrep-then-type. – dmckee Oct 1 '16 at 21:22

I think that you are best off by using some of the other data provided at the website that you are using although as pointed out in one of the comments information about the daughter is an important part of the jigsaw.

Betas from 60Co (5.2714 y 5)
Eb endpoint (keV)   Ib (%)      Decay mode
318,13              99.925 20   b-
665,26               0.022      b-
1491,38              0.057 20   b-

Gammas from 60Co (5.2714 y 5)
Eg (keV)     Ig (%)    Decay mode
346.93 7     0.0076 5     b-
826.06 3     0.0076 8     b-
1173.237 4  99.9736 7     b-
1332.501 5  99.9856 4     b-
2158.57 10   0.00111 18   b-
2505         2.0E-6 4     b-


The Q-value for the decay is given as 2823.95 keV.

From such data you can build up an energy level diagram and you from the data above you could complete the annotation for the left hand decays:

From the diagram you can see that the beta decays result in excited Nickel nuclei which then go to the ground state via the emission of gammas.
I would not worry about the gamma decays not being shown as 100%.
Overall the energy released whatever gamma decay modes actually occur is the same when the Nickel nucleus reaches its ground state.

• Thanks. My issue with that site is that I cannot just download all the data for all isotopes in bulk. I am trying to find a solution that uses data I can get easily. – nucleon Oct 1 '16 at 12:22