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$I_{131}\rightarrow \beta \text{ 333.8 KeV(7.27%)}\rightarrow \gamma\text{ 636.989 KeV(7.17%)}$

$I_{131}\rightarrow\beta\text{ 606.3KeV(89.9%)}\rightarrow\gamma\text{ 364.489KeV(81.7%)}$

(you can also check the them here: [Click Here])(

we can see there has 2 line of iodine 131 decay, there release different mean amount of energy.

  1. Normally,which radiation should we choose to measure the decay intensity? $\gamma$ or $\beta$?(such as when we measure the iodine from the air)

2.which decay line we should choose to detect?and why? enter image description here

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up vote 2 down vote accepted

There are actually quite a few more decay paths for $^{131}$I. In a practical situation, however, you don't select a particular decay to detect because your detector (unless you are using a single channel analyzer) will detect all of the radiation within it's range. In practice, you collect a gamma spectrum* from your sample and then analyze that spectrum to pick out what elements are there and in what ratios.

There are several programs (e.g. Genie 2000, GammaVision, PeakEasy) that will "automatically" analyze gamma spectra. Even with these programs, however, it is suprisingly easy to screw up the analysis. A well trained operator is by far the most significant factor in the quality of the analysis. I personally saw fresh data comming in from the Fukushima disaster and even in that situation many of the claims made about what was and wasn't present were questionable at best.

*select isotope 131 from I in the link for an example

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Ok,i understand so that mean for iodine,we detect all decay paths,and according to the result to analysis the original elements. – Mark0923 Oct 16 '12 at 12:47
Yes. This is the same process with any isotope, not just for Iodine. Actually, this is the same for any experimental science. Instruments are calibrated with well known standard to check the instrument; then the instrument is used to collect a signal from an unknown sample; then the sample signal is compared to theory and previous data to determine the nature of the sample. – AdamRedwine Oct 16 '12 at 18:07

In most applications the iodine is in some system that you're testing e.g. looking for leaks in a hydraulic system or measuring water circulation. In this case you look for the gamma radiation because the beta radiation is easily absorbed and usually doesn't make it out of the system. The gamma radiation is much more penetrating and can be detected at a distance.

In other circumstances beta absorption may not be a problem, e.g. measuring fallout from the Japanese nuclear incident, but you'd probably still use the the gamma radiation simply because that's what all the industrial detectors are calibrated for.

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if we use gamma and beta,respectively,can we get a same result? – Mark0923 Oct 16 '12 at 12:50
The trouble with beta radiation is that you get a spread of energies because the neutrino carries away some of the energy. If you're analysing a mixture of radionucleides you'd use the gamma radiation as it's more specific. If you just wanted to know how much iodine was present and there weren't any other radionucleides around and absorption of the electrons wasn't a problem then you could measure the beta radiation. – John Rennie Oct 16 '12 at 13:58 it,thank you~ – Mark0923 Oct 16 '12 at 14:11

As a rule of thumb you examine all the modes that

  1. you are sensitive to, and
  2. have enough counts to show above the background

in order to maximize the precision with which you know the answer and to insure that you have the isotope you think you have an are not seeing some some other decay that falls withing your detector resolution of one particular peak (i.e. you check the ratios between the modes, too).

The latter reason is especially important if you have modest detector resolution (as when using a NaI crystal of a liquid scintillator cell), the former when there is a significant amount of noise (say a lightly shielded germanium detetor being used to seek small activities).

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so that means,we choose γ because it can be easily identify?the resolution of the peak from γ is better than β? – Mark0923 Oct 17 '12 at 8:39

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