Luminescence Property of Ca_4Y_6Si_6O_(25)∶Eu~(2+) Green Phosphor

Ca4Y6Si6O25∶Eu2+ green phosphor was prepared by high temperature solid-state reaction.The XRD speetrum indicates that Ca4Y6Si6O25 has a hexagonal crystal structure with a space group of P63/m(176).The excitation and emission spectra of Ca4Y6Si6O25∶Eu2+ were measured when the mole fraction of Eu2+ was 20%.Its excitation spectrum is a broad band,from 350 nm to 450 nm,which matches with near ultraviolet LED chip.The emission spectrum of Ca4Y6Si6O25∶Eu2+ exhibits an asymmetry broad band with the peak center at 527 nm,and it is because that Eu2+ takes the place of Ca2+ and forms two different luminescence centers.The effect of Eu2+ concentration on Ca4(1-x)Y6Si6O25∶4xEu2+ intensity was also investigated.With the increase of Eu2+ concentration,the luminescence intensity of the green phosphor get to maximum and then decrease.The optimal mole fraction of Eu2+ is 20%(x=0.05).According to Dexter theory,concentration quenching can be attributed to quadrupole-quadrupole interaction.