Abstract
The first excited state (122–keV) of the even-even nucleus Sm152 was excited by the Coulomb excitation with 2-MeV protons. The angular distribution of the de-excitation radiation was investigated. The attenuation factor of the angular distribution G2∼0.6 was obtained using a samarium oxide target. The rotational change of the pattern in the angular distribution was examined under the condition of applying an external magnetic field vertical to the incident protons. The apparent g-factor for this state obtained was g(Sm152,2+)=+(0.36±0.16) n.m. A brief discussion about the effect of nuclear surroundings affecting the present determination was given.