The polarization of ${}_{}{}^{12}{}_{}{}^{}\mathrm{B}$ produced in the reaction ${}_{}{}^{100}{}_{}{}^{}\mathrm{Mo}$(${}_{}{}^{14}{}_{}{}^{}\mathrm{N}$, ${}_{}{}^{12}{}_{}{}^{}\mathrm{B}$)${}_{}{}^{102}{}_{}{}^{}\mathrm{Ru}$ is explained in a fully quantum mechanical way. It is found that recoil plays a decisive role.

• Received 2 August 1978

DOI:https://doi.org/10.1103/PhysRevLett.41.1770