He+2团簇的形成机理与结合能计算

当He+离子与He原子相互作用时, 由于一个电子往返运动于两核之间形成单电子键, 从而使He+与He结合成为具有较强键能的He+2.根据此形成机理, 采用简单的变分波函数, 计算了He+2基态的能量曲线.结果显示,当核间距为1.74 a 0时, 能量有一极小值-0.090 14 a.u.(以He++He能量为零起始计算).从而得到He+ 2离子结合能为0.090 14 a.u., 这与实验结果0.090 96 a.u.符合得相当好, 比有的理论计算值更接近实验结果.

The formation mechanism and the binding energy of the He+2 c luster

Authors think that when He ion interacts with He atom, because one electron goes back and forth between two nuclei, the He 2 can be formed with a strong binding energy. According to the formation mechanism, the energy curve for this structure has been calculated by using a simple variable wave function. The result shows that a minimal energy of -0.090 14 a.u. occurs at a separation of 1.74 a 0 between the two nuclei. So get the binding energy of 0.090 14 a.u. (start from the energy of He He). The binding energy agrees with the experiment result of 0.090 96 a.u. much better than some other results of calculation.