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I am going through An Introduction to Applied and Environmental Geophysics, a book by John Reynolds (2011), to better understand radiometrics and gamma-ray spectrometry. The author provides some tables about decaying as well as a plot of airborne gamma-ray spectrometry data.

Table 15.2A shows the Uranium decay series, while Table 15.2C shows the Thorium decay series. The plot shows the survey result of an airborne gamma-ray spectrometry.

As one can see in the plot, the author relates

  • the Potassium series with peak at 1,46 MeV with corresponding element 40K
  • the Uranium series (shown in Table 15.2A) with peak at 1,76 MeV with corresponding element 214Bi
  • the Thorium series (shown in Table 15.2C) with peak at 2,82 MeV with corresponding element 208Tl

The problem is: for both Uranium and Thorium decay series shown, the decay values presented at the tables differ from the peaks at the plot. I would expect to see the Uranium series peak with a value of 3.272 MeV (214Bi decay value in Table 15.2A) instead of 1,76 MeV shown in the plot.

The same for Thorium series: if the corresponding decay energy of 208Tl is 5.001 MeV accordingly to Table 15.2C, why does the plot show a corresponding value of 2.82 MeV?

Uranium decay series Thorium decay series airborne gamma-ray spectrometry

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I will first focus on the 214Bi decay. The 3.272 MeV is the total amount available for the β- decay, i.e. the difference between the ground states of 214Bi and 214Po. If you have a beta decay into the ground state of of 214Po, which happens in 19.9 % of cases (all data are from Firestone, R. B. et al., “Table of Isotopes”, 8th ed., 1998, p. 9211, in short “ToI”), there will be no gamma emission from 214Po, the 3.272 MeV are distributed between the electron and the electron antineutrino. However, in 16.9 % of cases, the decay leads to an excited state of 214Po with an energy of 1764.499 keV which can then deexcite among other possibilities by emission of a 1764.494 keV gamma photon (15.36 % of beta decays). The decay can produce a large number of gamma ray lines, but most happen with low probability. You can have a look at page 23 of this 214Bi table of radionuclides extract to get an impression.

In summary, the 3.272 MeV get distributed in varying proportion to the electron and the electron antineutrino from the beta decay, zero or more gamma-ray photons, and some small amount to nuclear recoil, etc.

In the case of 40K decaying via electron capture to 40Ar (10.72 % of cases), the total decay energy (Q-value) is 1504.9 keV. Most of the time (10.67 %), this goes to an excited state (1460.859 keV), which in turn gives rise to the 1460.830 keV gamma-ray line. The rest of the energy goes to the electron neutrino.

Similarly, for 208Tl, the 5000.9 keV is the Q-value of the β--decay. Most of the time (48.7 %), the decay proceeds via an excited state (3197.743 keV) of 208Pb, which deexcites via another excited state (2614.551 keV) where 99 % of decays also lead. This state the deexcites via the emission of a 2614.533 keV gamma ray photon. I believe the 2.82 MeV from Fig. 15.1 is a typo, since the reference in Reynolds’ book points to the article “Airborne Gamma-Ray Spectrometry Surveys”. There Fig. 4.4-1 is the same as Fig. 15.1 from the book, but in rather low resolution and with hard-to-read numbers. The text, however discusses the 2.62 MeV gamma ray line, which is very likely a rounded value of the 2614.533 keV from the “Table of Isotopes”.

The sum of the probabilities of different gamma ray lines from the same decay can be larger than 100 % if the deexcitation happens via a gamma-ray cascade, i.e. if the nucleus first transitions into an intermediate excited state and then again to the ground state or further intermediate states. This means that a single beta decay results in multiple gamma-ray emissions. This is illustrated in the Wikipedia article on Gamma rays using the example of the 60Co decay which results in two photons emitted in > 99.9 % of cases.

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  • $\begingroup$ First of all, thank you @Karsten Kretschemer. Secondly, your explanation makes a lot of sense for Bi. Why wouldn't it for Tl? I just can imagine that or either your explanation about Bi is a great coincidence (which I personally don't believe) or may there be some mistyped data? Can't see why. Third, why does the Gamma-ray probabilities (firstmost column to the right of energy column) from Bi sum up 100% (or seems so), while for Tl it surpasses 100%? $\endgroup$ – Leonardo Miquelutti Nov 25 '18 at 18:57
  • $\begingroup$ @LeonardoMiquelutti I have edited to hopefully address your comment $\endgroup$ – Karsten Kretschmer Nov 26 '18 at 10:41

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