ON THE ANGULAR DISTRIBUTION OF NEUTRONS IN C$sup 13$(/cap alpha/,n)O$sup 16$ REACTION

The angular distribution of neutrons in C/sup 13/( alpha ,n) were measured for four alpha energies, E/sub alpha / = 2.69, 2.83, 4.42, and 4.63 Mev. The finite O/sup 16/ nuclei are produced in their ground state. The character of the angular distribution changes slowly in relation to alpha energy. Moreover, the angular distributions have a sharply expressed forward and backward maximum. These factors lead to the assumption that the reaction occurs due to direct interaction with C/sup 12/ in which the alpha particle penetrates the nucleus, knocks out the neutron, and is absorbed by the C/sup 12/ field (forward pulses) or the neutron is liberated by the alpha particle analogous to a (d,n) reaction backward pulse). The angular distribution calculations and experimental data are plotted, considering the spin of C/sup 13/ = 1/2/sup -/, spin O/sup 16/ = 0/sup +/, spin He/sup 4/ = 0/sup +/, and the odd neutron in C/ sup 16/ is in the S state. In both processes the absorption moment of the motion is equal to zero. (R.V.J.)