【摘要】 采用高温固相法合成了Ba(1-x)SrxZr(BO3)2∶Eu3+系列样品,样品Ba(1-x)SrxZr(BO3)2∶Eu3+激发谱在130~170 nm和230 nm区域有两个很强的吸收带,位于130~170 nm的吸收带主要是硼酸盐基质的吸收;位于230 nm附近的吸收主要是Eu3+电荷转移态的吸收。当在样品中以Al部分取代Zr时,电荷转移态的吸收明显增强,并且Ba(1-x)SrxZr(BO3)2∶Eu3+发射强度也会明显增强;随着x的增大,硼酸盐基质的吸收强度减弱,基质吸收带的主峰值向低能方向移动了大约30 nm。样品Ba(1-x)SrxZr(BO3)2∶Eu3+在147nm激发下,发射出主峰值位于616 nm的强红光,对应Eu3+电偶极(5D0→7F2)跃迁发射。更多还原

【Abstract】 Powder samples of Ba_（1-x）Sr_xZr（BO₃）₂∶Eu³⁺（0≤x≤1） were prepared by the conventional solid state reaction method.The excitation spectrum shows a strong absorption in the VUV region with the broad absorption band edge at 130-170 nm due to the host and the f→d transition of Eu³⁺,and another absorption band of charge-transfer state（CTS） of Eu³⁺ in the region of 230 nm.The absorption of the charge-transfer band of Eu³⁺ can be increased by incorporating Al³⁺ ion into Zr⁴⁺ ion lattice.The luminescence of Ba_（1-x）Sr_xZr（BO₃）₂∶Eu³⁺ is also increased by incorporating Al³⁺ into the lattices.In Ba_（1-x）Sr_xZr（BO₃）₂∶Eu³⁺,the intensity of the absorption band（130-170 nm） decreases,and the band shifts about 30 nm toward lower energy with x increasing.The emission spectra revealed an intense and sharp（616 nm） red color emission from Eu ³⁺（（）⁵D₀→（）⁷F₂） transition under 147 nm VUV excitation.更多还原