Analysis of Luminescence Performance of New Red Phosphor K5Gd(MoO4)4∶xSm3+,yEu3+
A series of K5Gd(MoO4)4∶xSm3+(x=0～0.10)and K5Gd(MoO4)4∶0.04Sm3+,yEu3+(y=0.03～0.15)red phosphors were synthesized by high-temperature solid-phase method in this paper.The photoluminescence spectra,fluorescence emission spectra,and fluorescence quenching performance of the phosphor were analyzed by X-ray diffraction(XRD),fluorescence excitation spectra,thermal quenching analysis system,and steady-state transient spectrometer.Results show that the samples synthesized by doping Eu3+and Sm3+into K5Gd(MoO4)4 don't contain impurity phases,and the crystal structure remains unchanged.Under the excitation of 405 nm ultraviolet light,K5Gd(MoO4)4∶0.04Sm3+and K5Gd(MoO4)4∶0.04Sm3+,0.12Eu3+can emit excellent red light with the CIE coordinates of(0.6088,0.3904)and(0.637 3,0.359 2).In the K5Gd(MoO4)4∶xSm3+(x=0～0.10)samples,as the Sm3+doping concentration increases,the luminescence intensity of the phosphor first increases and then decreases,the optimal doping concentration is x=0.04.In the K5Gd(MoO4)4∶0.04Sm3+,yEu3+(y=0.03～0.15)samples,the luminescence intensity of Eu3+first increases and then decreases with increasing Eu3+doping concentration,and concentration quenching occur at y=0.12.When the temperature reaches 373 K,the fluorescence intensity of the K5Gd(MoO4)4∶0.04Sm3+phosphor sample is 94.69％of that at 293 K,while the fluorescence intensity of the K5Gd(MoO4)4∶0.04Sm3+,0.12Eu3+phosphor sample is 76.3％of that at 293 K,indicating their good thermal stability.The color coordinate graph shows that,doping of Eu3+leads to the color coordinate shifts slightly from the orange red region to the pure red region.Both K5Gd(MoO4)4∶xSm3+and K5Gd(MoO4)4∶0.04Sm3+,yEu3+phosphor samples have the potential to be used as red phosphors for white light LED.
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