Role of the Δ Resonance in the Population of a Four-Nucleon State in the ^{56}Fe→^{54}Fe Reaction at Relativistic Energies

The ^{54}Fe nucleus was populated from a ^{56}Fe beam impinging on a Be target with an energy of E/A=500  MeV. The internal decay via γ-ray emission of the 10^{+} metastable state was observed. As the structure of this isomeric state has to involve at least four unpaired nucleons, it cannot be populated in a simple two-neutron removal reaction from the ^{56}Fe ground state. The isomeric state was produced in the low-momentum (-energy) tail of the parallel momentum (energy) distribution of ^{54}Fe, suggesting that it was populated via the decay of the Δ^{0} resonance into a proton. This process allows the population of four-nucleon states, such as the observed isomer. Therefore, it is concluded that the observation of this 10^{+} metastable state in ^{54}Fe is a consequence of the quark structure of the nucleons.

Coulomb excitation of 104Sn and the strength of the 100Sn shell closure

A measurement of the reduced transition probability for the excitation of the ground state to the first 2+ state in 104Sn has been performed using relativistic Coulomb excitation at GSI. 104Sn is the lightest isotope in the Sn chain for which this quantity has been measured. The result is a key point in the discussion of the evolution of nuclear structure in the proximity of the doubly magic nucleus 100Sn. The value B(E2; 0+ → 2+) = 0.10(4) e2b2 is significantly lower than earlier results for 106Sn and heavier isotopes. The result is well reproduced by shell model predictions and therefore indicates a robust N = Z = 50 shell closure.