Deuteron Formation in the ReactionC12(e, e′d)BT=110

In the reaction ${}_{}{}^{12}{}_{}{}^{}\mathrm{C}(e, e^{\prime }d){}_{}{}^{10}{}_{}{}^{}\mathrm{B}$ the lowest-lying $T=1\mathrm{}$ state in ${}_{}{}^{10}{}_{}{}^{}\mathrm{B}$ is found to be as strongly excited as the $T=0\mathrm{}$ ground state of ${}_{}{}^{10}{}_{}{}^{}\mathrm{B}$, although the transition to the $T=1\mathrm{}$ state is isospin forbidden for direct deuteron knockout. A mechanism integration of a $p-n$ pair in a relative $T=1\mathrm{}$ state into a deuteron is proposed to explain this result. This new proposed mechanism is consistent with both the observed purely transverse character of the transition and the momentum-transfer dependence of the cross section.