【摘要】 在无任何保护性介质存在下以Ｎａ２ＳＯ３ 为除氧剂， ＫＩ为重原子微扰剂， 菲即能产生强而稳定的室温磷光（ＲＴＰ） 发射， λｅｘ／λｅｍ ＝ ２８３ｎｍ／４８２ ， ５０４ｎｍ ． 不同有机溶剂存在对其ＲＴＰ性质具有不同影响． １ ％ 乙腈存在时， 菲浓度在０－８ ～６－０ 和６－０ ～４０μｍｏｌ／Ｌ范围内分别与其ＮＰ－ ＲＴＰ 强度呈良好线性关系， 检出限为２－６ｎｍｏｌ／Ｌ．更多还原

【Abstract】 A strong and stable room temperature phosphorescence (RTP) signal of phenanthrine aqueous solution in the absence of a protecting medium can be induced only by using Na 2 SO 3 as deoxygenator and KI as heavy atom pertuber.The maximum phosphorescence intensity wavelenths are λ ex /λ em =283 nm/482,504 nm. It is also found that the kind and amount of organic solvent added to the luminescent system effects obviously the RTP properties. In the three organic solvents tested,effects of acetonitrile on NP-RTP emission of phenanthrine are less than those of alcohol or acetone. Only in the presence of 1% acetone,not only the RTP intensity was decreased about 70%. but also the exciting wavelength was shifted towards longer wavelenth.Under the present of 1% acetonitrile The RTP intensity is linear to phenanthrine concentration in the rage of 0 8～6 0 μmol/L and 6 0～40 μmol/L,respectively.The detection limit is 26 nmol/L.更多还原