Abstract

In this study, some of the fundamental aspects of neutralization of low-energy (10--300-eV) ${\mathrm{D}}^{+}$ ions have been addressed in comparison with that of ${\mathrm{He}}^{+}$ ions. To examine the role of the valence-band structure in determining neutralization rates for ${\mathrm{D}}^{+}$, a range of target materials has been studied (Mo, Ta, Ag, Pb, ${\mathrm{BaF}}_{2}$, CsCl, LiCl, AgCl, ${\mathrm{PbCl}}_{2}$, etc.). The surface peak of ${\mathrm{D}}^{+}$ is extremely small compared with that of ${\mathrm{He}}^{+}$ in scattering from metal surfaces, while the contrary is sometimes true in scattering from ionic compounds such as alkali-metal halides and alkaline-earth halides. It is found that the neutralization of ${\mathrm{D}}^{+}$ depends on the local electronic states of the target atom and occurs via the resonant neutralization by s- or d-band electrons of the metal atoms or via an energy-level-crossing mechanism between D 1s and the closed p band of the anions. The reason why the neutralization probability for ${\mathrm{D}}^{+}$ is very much enhanced compared with that for ${\mathrm{He}}^{+}$ is that the surface electronic state relevant to the resonant neutralization of ${\mathrm{D}}^{+}$ is not a level but a broad valence band (band effect). It is concluded that the smaller ionization energy of deuterium than helium results in strong coupling of the D 1s level with the valence band.

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