(p,t) Reaction on the Isotopes of Sm and Nd,N=88 and 90

Abstract

The ($p,t$) reaction on targets of ${\mathrm{Sm}}^{150,152}$ and ${\mathrm{Nd}}^{148,150}$ has been investigated using the 40-MeV beam of the University of Minnesota Linear Accelerator. Angular distributions have been measured for the lowlying levels in the range 5\ifmmode^\circ\else\textdegree\fi{} to 40\ifmmode^\circ\else\textdegree\fi{}. These nuclei are interesting in that they have $N=88 \mathrm{and} 90$, and span the rather sharp transition from a region of spherical to deformed nuclei. Three equally strong $L=0$ transitions are observed in the ${\mathrm{Sm}}^{152}(p,t){\mathrm{Sm}}^{150}$ spectrum corresponding to the excitation of ${0}^{+}$ states in ${\mathrm{Sm}}^{150}$ at 0,0.78, and 1.28 (\ifmmode\pm\else\textpm\fi{}0.05 MeV). The $L=0$ transition to a ${0}^{+}$ state at 1.28 MeV in ${\mathrm{Sm}}^{150}$ appears to correspond to the excitation of a deformed state in a nucleus with an approximately spherical ground state. One might expect the ($p,t$) reaction to excite such a state because the target nucleus, ${\mathrm{Sm}}^{152}$, is deformed. The 0.78-MeV group in the ${\mathrm{Sm}}^{152}(p,t){\mathrm{Sm}}^{150}$ spectrum probably corresponds to the excitation of the known ${0}^{+}$, "two-phonon," state at 0.743 MeV. In the ${\mathrm{Sm}}^{150}(p,t){\mathrm{Sm}}^{148}$ reaction, where both the target and residual nuclei have spherical ground states, the ground-state $L=0$ transition is the only state observed with significant strength below 1.45 MeV. A group at 1.45 MeV, which probably includes some $L=0$ strength, is observed with about $\frac{1}{3}$ the strength of the ground-state transition. The ${\mathrm{Nd}}^{150}$ and ${\mathrm{Nd}}^{148}(p,t)$ spectra are similar to the two Sm spectra with the same neutron number. Comparison is made with the existing ($t,p$) data on the same Sm nuclei.