The violation of the isospin selection rule has been studied in the reaction ${\mathrm{C}}^{12}(d,\ensuremath{\alpha}){\mathrm{B}}^{10}$ with deuteron energies between 9 and 12.5 MeV. The differential and total cross sections of the isospin-forbidden transition to the first $T=1$ state in ${\mathrm{B}}^{10}$ (the ${J}^{\ensuremath{\pi}}={0}^{+}$ state at 1.74-MeV excitation) have been compared with the cross sections of the isospin-allowed transitions to the ground state (${J}^{\ensuremath{\pi}}={3}^{+}$, $T=0$), the first excited state (${J}^{\ensuremath{\pi}}={1}^{+}$, $T=0$ at 0.72 MeV), and the third excited state (${J}^{\ensuremath{\pi}}={1}^{+}$, $T=0$ at 2.14 MeV). Mostly the intensity of the $T=1$ alpha group is less than 1% of the yield of the $\ensuremath{\alpha}$ groups leading to the neighboring $T=0$ states. This reduction is due not only to isospin forbiddenness but also to angular-momentum and parity selection rules which apply in this particular ($d,\ensuremath{\alpha}$) reaction for which both the initial and the final state have ${J}^{\ensuremath{\pi}}={0}^{+}$. These weighting factors have been calculated by use of the statistical theory of nuclear reactions. After these factors have been applied, the $T=1$ alpha group has an intensity of about 10% relative to the other three $T=0$ transitions at a deuteron energy of 9 MeV and 1-2% at an energy of 11 MeV. The small yield is ascribed to the isospin selection rules that to some extent govern this $T=1$ transition. In the energy range from 9 to 11 MeV, the angular distribution of the $T=1$ state stays fairly constant and is nearly symmetric around 90\ifmmode^\circ\else\textdegree\fi{}. The yield decreases steadily. At deuteron energies higher than 11.5 MeV, the angular distribution changes drastically and becomes strongly forward peaked and asymmetric around 90\ifmmode^\circ\else\textdegree\fi{}, and the total yield increases slightly. We assume that this behavior indicates a direct-interaction mechanism in which the process of mixing the isospins takes place at the surface of the nucleus. Coulomb excitation during the process of $d$ capture or $\ensuremath{\alpha}$ emission might be responsible for the isospin violation at these higher deuteron energies.
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