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卤代烷RX的第一电离能Ip的变化规律可用如下方程表示：Ip（eV）＝3.3380＋0．7344Ep（X）＋2.7424[（1／ax）qx]－1．4302PEI其中，Ep（X）、αx分别为卤素原子最外层P电子的电离能和原子极化度，qx是卤代烷RX分子中卤原子X所带的部分电荷，PEI是RX分子中烷基R的极化效应指数．研究结果表明用上式估算卤代烷的第一电离能与实验值符合的比较好．
The variation of the first ionization energy Ip of the alkyl halide RX can be expressed by the following equation: Ip (eV) = 3.3380 + 0.7344Ep (X) +2.7424 [(1 / ax) qx] -1.4302PEI where Ep αx are the ionization energies and atomic polarizabilities of the outermost P electron of the halogen atom, qx is the partial charge carried by the halogen atom X in the alkyl halide molecule, and PEI is the polarization index of the alkyl group R in the RX molecule. The results show that using the above formula to estimate the first ionization energy of alkyl halides is in good agreement with the experimental data.