Theoretical study of $ \rm{Nb} $ isotope productions by muon capture reaction on $ {}^{100} \rm{Mo} $

Abstract

The isotope $ {}^{99} \rm{Mo} $, the generator of $ {}^{99m} \rm{Tc} $ used for diagnostic imaging, is supplied by extracting from fission fragments of highly enriched uranium in reactors. However, a reactor-free production method of $ {}^{99} \rm{Mo} $ is searched over the world from the point of view of nuclear proliferation. Recently, $ {}^{99} \rm{Mo} $ production through a muon capture reaction was proposed and it was found that about $ 50 \, \% $ of $ {}^{100} \rm{Mo} $ turned into $ {}^{99} \rm{Mo} $ through $ {}^{100} \rm{Mo} \left( μ^-, n \right) $ reaction [arXiv:1908.08166]. However, the detailed physical process of the muon capture reaction is not completely understood. We, therefore, study the muon capture reaction of $ ^{100} \rm{Mo} $ by a theoretical approach. We used the proton-neutron QRPA to calculate the muon capture rate. The muon wave function is calculated with considering the electronic distribution of the atom and the nuclear charge distribution. The particle evaporation process from the daughter nucleus is calculated by a statistical model. From the model calculation, about $ 38 \, \% $ of $ {}^{100} \rm{Mo} $ is converted to $ {}^{99} \rm{Mo} $ through the muon capture reaction, which is in a reasonable agreement with the experimental data. It is revealed that negative parity states, especially $ 1^- $ state, play an important role in $ {}^{100} \rm{Mo} \left( μ^-, n \right) {}^{99} \rm{Nb} $. The feasibility of $ {}^{99} \rm{Mo} $ production by the muon capture reaction is also discussed. Isotope production by the muon capture reaction strongly depends on the nuclear structure.