Identification ofHo148andHo149

The new nuclides, ${}_{}{}^{148}{}_{}{}^{}\mathrm{Ho}$ and ${}_{}{}^{149}{}_{}{}^{}\mathrm{Ho}$, were identified in ${}_{}{}^{10}{}_{}{}^{}\mathrm{B}$ bombardments of ${}_{}{}^{144}{}_{}{}^{}\mathrm{Sm}$. A gas-jet system was used to transport product nuclei to a shielded area suitable for $\gamma$-ray measurements. The 21-s isotope, ${}_{}{}^{149}{}_{}{}^{}\mathrm{Ho}$, was found to decay primarily to a ${}_{}{}^{149}{}_{}{}^{}\mathrm{Dy}$ level at 1091 keV and less intensely to the $i_{\frac{13}{2}}1073$-keV state observed in a previous in-beam $\gamma$-ray study. This radioactivity, in analogy to neighboring odd-$A$ terbium and holmium isotopes, probably represents the $h_{\frac{11}{2}}$ proton orbital in ${}_{}{}^{149}{}_{}{}^{}\mathrm{Ho}$ which decays via an allowed $\beta$ transition to the 1091-keV level. The latter, is therefore, proposed to be the $h_{\frac{9}{2}}$ neutron state in ${}_{}{}^{149}{}_{}{}^{}\mathrm{Dy}$. The 9-s isotope, ${}_{}{}^{148}{}_{}{}^{}\mathrm{Ho}$, was identified mainly through a 1688-keV $\gamma$-ray which: (1)was in coincidence with dysprosium $K$ x rays, (2)increased dramatically in intensity when the ${}_{}{}^{10}{}_{}{}^{}\mathrm{B}$ bombarding energy was raised from 85 to 96 MeV, and (3)remained constant (over the same range) in intensity relative to that of the 620-keV $\gamma$ ray known to belong to ${}_{}{}^{148}{}_{}{}^{}\mathrm{Dy}$ decay. On the basis of systematics for low-lying levels in even-even $N=82\mathrm{}$ nuclei, the 1688-keV transition is though to deexcite the first-excited state in ${}_{}{}^{148}{}_{}{}^{}\mathrm{Dy}$.