Angular momenta of even-even fragments in the neutronless fission of<mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" display="inline"><mml:mrow><mml:msup><mml:mrow/><mml:mrow><mml:mn>252</mml:mn></mml:mrow></mml:msup></mml:mrow><mml:mi mathvariant="normal">Cf</mml:mi></mml:math>

The recent advent of experimental techniques in which the dynamical characteristics of fission fragments are determined more accurately prompted us to investigate the angular momentum acquired by fragments in a model which describes the cold (neutronless) fission of ${}^{252}\mathrm{Cf}$ as the decay of a giant nuclear molecule. The molecular configuration is a consequence of the interplay between the attractive nuclear part and the repulsive Coulomb+nuclear forces. The basic idea of the present approach is to separate the radial (fission) modes describing the decay of the molecule from the modes associated to transversal vibrations (bending) of the fragments. The distance between the centers of the two fragments is fixed by the requirement that the energy released in the fission reaction Q equals the sum of quantum zero energies of radial and transversal modes and the total excitation energy ${E}^{*}.$ Using a semiclassical coupled channel formalism we computed the additional angular momenta acquired by the fragments during their postscission motion, and found that the Coulomb excitation accounts for less than 10% of the final spins.

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