Some properties of the 7.1-MeV level of ${\mathrm{Na}}^{23}$ have been studied using the resonance fluorescence technique with the ${\mathrm{F}}^{19}(p, \alpha ){\mathrm{O}}^{16*}$ reaction serving as the source of Doppler-broadened radiation. By comparison with the (7.115±0.003)-MeV level of ${\mathrm{O}}^{16}$ the energy was determined to be 7.10±0.02) MeV. The angular distribution of the scattered radiation is $W\left(\theta \right)=1+(0.47\mathrm{}\pm 0.09)P_2+(0.32\mathrm{}\pm 0.03)P_4$, which is consistent only with $\frac{5}{2}$ for the spin. The values allowed for the quadrupole-dipole amplitude ratio, $\delta$, are $-1.8>\delta >-2.2\mathrm{}$ and $2.5<\mathrm{}\delta <3.0\mathrm{}$. Using this spin value and the results of the self-absorption measurement, the partial width for the ground-state transition becomes equal to ${\Gamma }_0=(0.62\mathrm{}\pm 0.07)$ eV. With the observed branching of 63% to the ground state, the total width becomes $\Gamma =(0.98\mathrm{}\pm 0.12)$ eV. This large value of the width combined with the large value of $\delta$ defines the parity of the state as positive.

• Received 20 July 1964

DOI:https://doi.org/10.1103/PhysRev.136.B1355