ArCO2 IonizationPhysics

[math]\bar{E}=[/math] 27 eV = average energy to ionize and electron in an Argon Atom

[math]\bar{N_{PE}} = \frac{E_{\gamma}}{27 eV} =[/math] average number of photoelectrons produced

Quenching Gas:

1.) reduces the influence of the positive ions creates on the photoelectron signal: The excited Ar+ atoms emit photon in the UV range which are absorbed by the quenching gas

2.)Collisions with the quenching gas will neutralize the Ar+ ions. When the quenched gas, having an electron remove by the Ar+ collision, reaches the cathode and collects an electron, most of the energy goes into dissociation of the Quench gas.

If the quech gas is CH4 then

CH4+ [math]\Rightarrow[/math] H2 + CH2

Argon Escape peak

You need 3.2 keV to ionize a K-shell electron in Argon. If your incident ionizing particle (Photon or electron) has more than that energy then it is possible to excite Argon so it becomes a source of photons during the ionization process. If that photon ESCAPEs the detector without causing ionization, then your signal will be contain less ionized electrons.