Quenching of the2S01Metastable State of Helium by an Electric Field

Abstract

A time-of-flight technique has been used to measure the quenching of the $2^{1}S_{0}$ metastable state of helium by a static electric field. Neutral ground-state helium atoms effusing from a cooled source slit are immediately excited to the $2^{1}S_{0}$ and $2^{3}S_{1}$ metastable states by a pulse of antiparallel magnetically focused electrons. The metastable beam is collimated before passing through a uniform electric field region 0.5 m long and is then preferentially detected at the end of the time-of-flight region, 1.825 m from the electron gun. The time-of-flight distribution for the $2^{1}S_{0}$ state is separated from that of the $2^{3}S_{1}$ state by illuminating the beam before the electric field region with an rf-discharge helium lamp. The $2^{1}S_{0}$ state is quenched by resonant absorption of a 20 581-\AA{} photon, raising the atom to the $2^{1}P_{1}$ state, which then decays to the $^{1}S_{0}$ ground state; the $2^{3}S_{1}$ state remains unaffected because it is the ground state for the triplet system. The $2^{1}S_{0}$ time-of-flight distribution is therefore obtained from the difference between the full beam and the quenched beam. The number of $2^{1}S_{0}$ atoms arriving at the detector in specific velocity intervals with the electric field off is compared to the number with the field on to determine the quenching rate ($=k{E}^{2}$). The result for the quenching constant $k$ for both ${\mathrm{He}}^{3}$ and ${\mathrm{He}}^{4}$ with $E$ in kV/cm is $k=0.933\ifmmode\pm\else\textpm\fi{}0.005$; this value is in good agreement with theory and with an earlier less accurate experiment. The error in the present experiment arises from a 0.5% uncertainty in the effective length of the electric field region.