The location of the ground state of ${\mathrm{Er}}^{3+}$ in a ${\mathrm{Y}}_2{\mathrm{O}}_3$ single-crystal fiber is determined using two-step photoconductivity. The ground state of ${\mathrm{Er}}^{3+}$ ${(}^4{I}_{15/2})$ is located 1.0 eV (±0.3 eV) below the top of the valence band. The second step photocurrent results from excitation of a $4f$ electron from the ${}^4{S}_{3/2}$ excited state to the $5d$ state followed by photoionization into the conduction band.

• Received 1 March 2002

DOI:https://doi.org/10.1103/PhysRevB.65.235116