N14(p,p′)N14reaction between 25 and 40 MeV and the tensor part of the effective interaction

Differential cross sections for the scattering of protons from ${}_{}{}^{14}{}_{}{}^{}\mathrm{N}$, leaving this nucleus in its ground, 2.31- or 3.95-MeV states, were measured at bombarding energies of 29.8, 36.6, and 40.0 MeV. In addition, cross sections for the next ten excited states with $E_x<8.7\mathrm{}$ MeV were obtained at 29.8 MeV. Emphasis was placed on obtaining reliable results for the transition to the weakly excited $J^{\pi }=0^{+}$, $T=1\mathrm{}$ state at 2.31 MeV, which is unusually favorable for studies of the tensor part of the effective interaction. The present data for the 2.31-MeV state, together with results from the literature at $E_p=24.8\mathrm{}$ MeV, were analyzed using a microscopic model distorted-wave approximation; contributions from the knock-on exchange amplitude and from central, tensor, and spin-orbit two-body forces were included. The fits obtained were satisfactory at 24.8 and 29.8 MeV, but failed to reproduce a peak in the cross section which appeared near 80° at the higher bombarding energies. Strengths of the tensor force were extracted and were found to be 20-75% larger than estimates based on the one-pion-exchange potential. Available evidence on the strength of the tensor force is summarized. Of the other observed states, those whose structure is dominated by $1p$-shell orbitals [at 3.95 MeV ($1^{+}$, $T=0\mathrm{}$) and at 7.03 MeV ($2^{+}$, $T=0\mathrm{}$)] were compared with microscopic distorted-wave approximation calculations using an empirically derived central force. Cross section enhancement factors were calculated in an effective charge approximation. The resulting cross sections have roughly the correct magnitude, but the shapes are only qualitatively correct. A systematic optical model analysis was carried out to provide optical model parameters for use in the distorted-wave approximation calculations. The rate of change of the real potential with bombarding energy was found to be $\frac{d{V}_R}{d{E}_p}=-0.50\mathrm{}$.