# Study of mass yield and kinetic energy distribution of fission fragments from fission of the excited compound nucleus 254Fm produced in fusion reaction

Research paper by H. Eslamizadeh, M. Soltani

Indexed on: 04 Oct '18Published on: 05 Oct '18Published in: The European Physical Journal A

#### Abstract

A stochastic approach based on four-dimensional Langevin fission dynamics has been used to calculate the anisotropy of fission fragment angular distribution, the mass and energy distribution of fission fragments, the yield and variance of the mass distribution of fission fragments, the fission cross section and the pre-scission neutron multiplicity for the compound nucleus 254Fm produced in the reaction 16O + 238U. In the dynamical calculations dissipation coefficient of K , $$\gamma_k$$ , was considered as a free parameter and its magnitude inferred by fitting measured data on the anisotropy of fission fragments angular distribution. It was shown that the results of the anisotropy of fission fragments angular distribution for the compound nucleus 254Fm are in good agreement with the experimental data by using values of dissipation coefficient of K, equal to $$\gamma_{k}=(0.076-0.088)$$ (MeV zs)-1/2. We also calculated the above-mentioned experimental data for the compound nucleus 254Fm by using $$\gamma_{k}=(0.076-0.088)$$ (MeV zs)-1/2 and results of the calculations compared with the experimental data. A comparison of the theoretical results with the experimental data showed that the results of calculations are in good agreement with the experimental data, although at high energy the difference between the results of calculations with the experimental data for the variance of the mass distribution of fission fragments and the pre-scission neutron multiplicity is slightly high. It was also shown that at high energy the variance of the mass distribution of fission fragments and the pre-scission neutron multiplicity can be reproduced for the compound nucleus 254Fm by using values of dissipation coefficient of K equal to $$\gamma_{k}=0.098$$ (MeV zs)-1/2 and $$\gamma_{k}=0.065$$ (MeV zs)-1/2, respectively.