Isotopic spin selection rules XI: The 8.06 and 6.23 MeV states of14N

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Abstract
The decay scheme of the 8.06 MeV 1−T =1 level of 14N is reinvestigated using the reaction 13C(p, γ)14N at the 550 KeV resonance. The relative strength of the isotopic spin forbidden transition to the first excited state is 0.022 corresponding to α1 2(0)8.06∼0.061. Owing to the great strength to it of the transition from the 8.62 MeV O+ state it seems that the 6.23 MeV T = O state must be 1−. The decay of this state has been determined; it has been reached by this El transition from the 1.16 MeV resonance. The strength of the isotopic spin forbidden transition to the ground relative to the allowed transition to the first excited state is 0.31 corresponding to α0 2(1)6.23∼0.078.The closeness of these two isotopic spin impurities suggests that these two states may be each other's chief contaminators and in this Sense the forbidden transitions from the one is the same radiation as the allowed transition from the other and vice versa. If this suggestion is correct H C∼0.45 Mev which large value suggests some close relationship between the two states. In unravelling the decay schemes subsidiary pieces of information are obtained; in particular it seems that the 5.69 MeV state is very probably 1+.