Quenching of theHeμ+(2s) atom

Quenching of the metastable 2s state of the ${\mathrm{He}}_{\mathrm{\mu }}$ ${\mathrm{}}^{+}$ atom in helium gas is discussed. The first part of the discussion, which is devoted entirely to processes occurring after the ${\mathrm{He}}_{\mathrm{\mu }}$ ${\mathrm{}}^{+}$ has become bound to one or more ordinary helium atoms, is based partly on Cohen’s calculations of rates of vibrational quenching and partly on estimates obtained in the present paper of rates of Burbidge–de Borde quenching and Ruderman quenching. It is concluded that Burbidge–de Borde quenching or Ruderman quenching, or both, are likely to be more effective than Cohen quenching if the vibrational level of the bound system is low. A recent experiment by von Arb et al. is then analyzed in the light of this conclusion. The analysis is based on the reported absence, or near absence, of Auger electrons accompanying 2s quenching. While it is agreed that the Cohen mechanism occurring in the molecular ion ${\mathrm{HeHe}}_{\mathrm{\mu }}$ ${\mathrm{}}^{+}$ remains the most likely explanation of the experiment, it is concluded that the quenching occurs in comparatively high levels. It is then argued that this conclusion is in accord with some theoretical investigations of three-body association reactions and also with some elementary considerations regarding the relaxation of highly excited diatomic molecules, and it is further concluded that the quenching is most likely to occur in states with very low rotational quantum number and vibrational quantum number 8≤v≤14.