How is iodine-131 produced in nuclear reactors? I coulnd't find clear explanations about the production of Iodine-131 in nuclear reactors based on U-235. 


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*Can it be derived directly from U-235 ? In this case : U-235 + neutron -> I-131 + what ??? 

*Or is it derived indirectly from U-235 ? In this case : U-235 + neutron -> X ? -> I-131 (what could be X? And by which process? neutron impact on X or beta-decay of X?) 
Thanks for any information.
 A: It is a fission product, one of many nuclides: http://hyperphysics.phy-astr.gsu.edu/hbase/NucEne/fisfrag.html
As uranium has Z=92 and iodine Z=53, the other piece could be yttrium (Z=39). But total Z may change because of all the beta-decays. The iodine isotope is likely a decay product of short-lived fission products like In-131, Sn-131, Sb-131, Te-131.
A: Uranium only serves as a neutron source. Yes, the decay reaction produces I-131, but it's in the radioactive waste, and it would be a pain to extract it. Your typical reactor fuel is left in for many years, during which quite a bit of the produced I-131 would decay again. Plus, the waste is quite radioactive, making it inconvenient to handle. And you get a lot of I-129 that you can't chemically separate.
Actual production is based on a convenient fact: Natural Tellurium can be radiated with neutrons from U-235 fission. This will produce a wide variation of isotopes, but if you give the short-lived isotopes time to decay (~few hours), let the Xenon escape, and chemically remove the remaining Tellurium, you'll find that the remaining mixture is very rich in I-131.
A: Most radiopharmaceuticals are accelerator produced. Less costly than building a reactor. Accelerators offer a lower radiation risk to workers and essentially zero risk to the public.
