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由瞬时碰撞模型出发,给出一种计算A+BC→AB+C型反应产物转动取向的方法。研究结果表明,产物转动角动量相对于反应物相对速度的取向由排斥能R释放后A、B原子动量大小之比唯一确定,比值越大,产物的转动取向越强。进一步得出,排斥能R增加,或C原子质量增加,都使产物转动取向减弱,而A原子速度增加使产物转动取向加强。此模型适用于碱金属及碱土金属原子与氧化剂分子的反应,计算结果和一些实验结果作了比较,二者符合较好。
Based on the instantaneous collision model, a method for calculating the rotational orientation of A + BC → AB + C reaction products is given. The results show that the ratio of the rotational angular momentum relative to the reactant relative velocity is only determined by the ratio of the momentum of the A and B atoms released by the repulsion energy R, The stronger the orientation of rotation, the stronger the repulsion energy R, or the increase of the mass of C atom, the weakened rotation orientation of the product, while the increase of the A atom speed enhances the rotation orientation of the product. This model is suitable for the reaction between alkali metal and alkaline earth metal atoms and oxidant Molecular response, calculated results and some experimental results were compared, both in good agreement.