Interpretation of the 21.2 eV photoelectron spectrum of atomic Ba

Abstract

The recent observation by other workers of an appreciable double ionization rate for photoionization of atomic barium using 21.2 eV photons is explained. The phenomenon can be explained as a two-step autoionization process in which an excited state of Ba, identified as 5p⁵5d¹P6s², first autoionizes into an excited state of Ba⁺ above its first ionization limit and a free electron. Subsequently this intermediate state autoionizes into the Ba²⁺ ground state and a second free electron. Three intermediate states have been observed and their identity is discussed.