采用高温固相法合成Sr5Si2O7Cl4:Eu^2+材料。在370nm激发下,Sr5Si2O7Cl4:Eu^2+材料呈现峰值位于490nm的非对称发射,对其进行Gaussian曲线拟合,得到峰值位于485nm和520nm两个明显的发射峰。利用van Uitert公式讨论Eu^2+在基质中的晶格环境和发光特性,确定晶体中有蓝和黄绿两种发光中心:485nm发射来源于七配位的Eu^2+中心发射;520nm长波发射与杂质束缚激子有关。进一步研究Sr5-xCaxSi2O7Cl4:Eu^2+材料的发光性质,当0〈x〈0.5,Ca^2+固溶于Sr5Si2O7Cl4基质晶格,Eu^2+占据七配位Sr^2+格位,晶体主要产生蓝色中心的蓝绿色发射;当0.5〈x〈2,Ca^2+掺量增加使晶胞参数变小,晶格中的杂质束缚激子态的束缚增强,Eu^2+处于杂质束缚激子中心所形成的激发态能量进一步降低,发射位于长波段,Sr5-xCaxSi2O7Cl4:Eu^2+主要产生黄绿色中心的黄绿色发射。
Phosphor Sr5Si2O7Cl4:Eu^2+was synthesized using the high temperature solid state method.Under the 370 nm excitation,phosphor Sr5Si2O7Cl4:Eu^2+shows one unsymmetrical emission at 490 nm.Two obvious emission peaks at 485 nm and 520 nm can be obtained through the Gaussian curve simulation.According to the van Uitert experimental formula,luminescent characteristics and crystal-lattice environment of Eu^2+ in Sr5Si2O7Cl4 are discussed.The two emission centers exist in the crystal,viz,blue and greenishyellow centers are verified.The 485 nm emission band originates from the Eu^2+center which inhabits the seven compounds,and the 520 nm emission band depends on the impurity-trapped exciton.Under the condition of doping Ca^2 +,the luminescent characteristics of the phosphor are investigated.When the doping amount of calcium(x) varies between 0 and 0.5,Ca^2 +are embedded in the host lattice of Sr5Si2O7Cl4,and the blue emission originates from Eu ^2+ which substitutes seven compounds Sr^ 2+sites,and the phosphor emits a blue-green light.Under the condition of 0.5〈x〈2,the parameters of the crystal system decreases with Ca^2 +content is increased,the impurity-trapped exciton is also strengthened,and the excitation state energy of Eu^2+which exists in the center of impurity-trapped exciton will decrease.Hence,a long wavelength emission appears,and emits a greenish-yellow luminescence.