A large isotopic-spin violation was observed in the ${\mathrm{Ne}}^{20}(d, \alpha ){\mathrm{F}}^{18}$ reaction. The yield to the second $T=1\mathrm{}$ level (3.063 MeV) in ${\mathrm{F}}^{18}$ was measured and compared with the yield to an adjacent (3.133 MeV) $T=0\mathrm{}$ level. The ratio of the yields to these two levels, averaged over angles from 50° to 140°, for an input energy of 4.00 MeV, was found to be 0.40. A value of approximately 1 was found for this ratio using the ${\mathrm{O}}^{16}(\mathrm{He}, p){\mathrm{F}}^{18}$ reaction. The isotopic-spin selection-rule breakdown is attributed to Coulomb mixing of states in the compound nucleus ${\mathrm{Na}}^{22}$ and in the residual nucleus ${\mathrm{F}}^{18}$. Angular distributions of 11 proton groups from the ${\mathrm{O}}^{16}({\mathrm{He}}^3, p){\mathrm{F}}^{18}$ reaction and of seven alpha groups from the ${\mathrm{Ne}}^{20}(d, \alpha ){\mathrm{F}}^{18}$ reaction are presented.

• Received 15 June 1964

DOI:https://doi.org/10.1103/PhysRev.136.B399