Kinetics of Electron and Ion Reactions during the γ Radiolysis of Liquid Neopentane; Dependence of Charge Scavenging Kinetics on Hydrocarbon Molecular Structure

Abstract

The large free ion yield in neopentane has been confirmed by chemical measurement: for nitrous oxide solutions at 23°C $\mathrm{g(}{\mathrm{N}}_2{)}_{\mathrm{fi}}\text{\hspace{0.17em}=\hspace{0.17em}0.95}$ and $\mathrm{g(}{\mathrm{N}}_2{)}_{\max }\text{\hspace{0.17em}≈\hspace{0.17em}4.7}$ . The average number of nitrogen molecules formed per electron scavenged appears to be 1.1 ± 0.1 at all nitrous oxide concentrations. Thus $g_{\mathrm{fi}}\text{\hspace{0.17em}=\hspace{0.17em}0.9}$ in neopentane, which is sevenfold larger than that in n‐hexane. The median range of the secondary electrons in liquid neopentane is about fourfold greater than that in n‐hexane. The greater range in neopentane correlates with the more sphere‐like shape of the molecules. In spite of the larger median electron range in neopentane, the ease with which electrons and ions can be scavenged in spurs in that liquid is similar to that in hydrocarbons with distinctly nonspherical molecules. Furthermore, the ratio of efficiences of electron to positive ion scavenging ${\mathrm{(\beta }}_{-}{\mathrm{\hspace{0.17em}/\hspace{0.17em}\beta }}_{+})$ is 17 in neopentane, similar to the value 13 in the “nonspherical” methylcyclopentane. During the radiolysis of pure neopentane two types of ions are formed: $\mathrm{G(}{\mathrm{PH}}^{+}\text{)\hspace{0.17em}=\hspace{0.17em}1.7}$ and $\mathrm{G(}{\mathrm{N}}^{+}\text{)\hspace{0.17em}=\hspace{0.17em}3.0}$ . PH⁺ yields hydrogen when neutralized by an electron, but not when neutralized by a massive negative ion. PH⁺ can also donate a proton to ammonia. N⁺ does not yield hydrogen upon neutralization, nor does it donate a proton to ammonia. In pure neopentane $\mathrm{G(}{\mathrm{H}}_2\text{)\hspace{0.17em}=\hspace{0.17em}2.5\hspace{0.17em}±\hspace{0.17em}0.1}$ and $\mathrm{G(}{\mathrm{CH}}_4\text{)\hspace{0.17em}=\hspace{0.17em}3.9\hspace{0.17em}±\hspace{0.17em}0.2}$ at a dose of 1.5 × 10¹⁸ eV/g. Scavenging the free ions by either N₂O or ND₃ reduces $\mathrm{g(}{\mathrm{H}}_2)$ by about 0.35, but has no effect on the methane yield.