Abstract

Abstract It is pointed out that the 8·06 mem (1-) state of 14N is of T=l, corresponding with the 6·10 mev state of 14C. The electric dipole transition to the 2·31 (0+) T=l state of 14N is therefore forbidden by the isotopic spin selection rules. It is shown that the strength of this forbidden electric dipole transition is less than 0·7% of that of the allowed electric dipole transition to the (1+) T=0 ground state of 14N ; this implies that the contamination of the 8·06 mev state by a state of T=0 is less than 2% in intensity.

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