Abstract
A six band rotation-particle coupling calculation is done for the $\frac{11}{{2}^{\ensuremath{-}}}$ states in $^{147}\mathrm{Sm}$. An entirely different distribution of the band energies is found as compared with the recently reported results from ($^{3}\mathrm{He}$, $\ensuremath{\alpha}$) reactions.NUCLEAR STRUCTURE Rotation-particle calculation for $1{h}_{\frac{11}{2}}$ neutron-hole states in $^{147}\mathrm{Sm}$.
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