Proton Polarization in theSr88(d,p)Sr89*(1.05-MeV) Reaction

The polarization of protons from the ${\mathrm{Sr}}^{88}\mathrm{}(d,p)\mathrm{}{\mathrm{Sr}}^{89*}$ reaction exciting the 1.05-MeV state in ${\mathrm{Sr}}^{89}$ has been measured at 14 laboratory angles from 12 to 131° using a deuteron bombarding energy of 11 MeV. In this reaction, which involves no orbital-angular-momentum transfer by the captured neutron, spin-orbit terms in the optical-model potentials describing the interactions of the incident and exit particles are considered to be responsible for the polarization. The measurements were made using a heavy-particle magnetic spectrometer to select the desired proton group and focus it on the carbon second scatterer. This scatterer was viewed to the "right" and the "left" by a pair of scintillation counters set at a mean laboratory angle of 40°. The sign of the polarization seems to correspond to the sign of the slope of the angular distribution, being negative at the forward angles with apparent sign changes at the subsequent minima and maxima. This behavior is in approximate agreement with the "derivative rule" for stripping reactions with $l_n=0\mathrm{}$ orbital-angular-momentum transfer. The polarization changes most rapidly near the stripping minima and becomes largest at the backward angles (-28% at 95°).