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Reply To: nuclear physics decay question
- Member28/11/2021 at 3:20 pm
radio-chemistry, an interesting topic and what got me interested in NDT 35 years ago. But the isotope you refer to is not one usually associated with NDT!
Some background for those not familiar;
Bromine has 2 stable isotopes, Br79 and Br81.
Br80 is one of several unstable isotopes of Bromine.
Also, Bromine 80 is a nuclear isomer. A nuclear isomer has the same nuclear mass and same atomic number but different radioactive properties. In this case the isomers are 35Br80 and 35Br80m (where 35 is the Z number or atomic number and 80 the atomic mass and 35 would be a subscript and 80 and the letter m are superscript).
Wikipedia describes nuclear isomers as “…a metastable state of an atomic nucleus caused by the excitation of one or more of its nucleons. A nuclear isomer occupies a higher energy state than the corresponding non-excited nucleus, called the ground state. Eventually, the nuclear isomer will release the extra energy and decay into the ground state,…”,.
In my old school text “Radio and Nuclear Chemistry” by Friedlander Kennedy and Miller (Wiley 1964) they note that the unstable isotope of Br80 decays by several decay methods including Beta negative with associated gamma emission, Beta positive (positron), and Electron Capture, and it has a short half-life of only 17.6 minutes.
Bromine 80m on the other hand has a half-life of 4.5 hours making it “metastable”. It decays by “Isomeric Transition (IT) via internal-coversion electrons.
Since my reference was quite old I decided to check other locations available online and confirmed that the 80m mode was indeed the metastable version and did not decay by beta negative or beta positive modes. Instead it is the less stable version with a 17.6minute half-life that decays with the beta negative or beta positive modes.
You asked WHY Bromine80 decays with 2 options; i.e. both beta negative and positron (beta positive). Generally it is considered that beta negative decay dominates on the neutron excess side of the mass-parabola and positron decay on the proton rich side of the stable isobar. The actual route of decay is related to a variety of selection rules in quantum mechanics. There are “allowed” and forbidden” transitions to de-excite to a more stable state. However, “forbidden” does not mean it is in fact not allowed, but instead exists with only a low probability.
Details of the classroom lectures on quantum mechanics describing the transitions are now too long ago in my past. I suggest you get a copy of Friedlander or some similar text to fill in the details (they are not trivial).