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 3.203 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 (X-rays) during the ionization process. The Ar-Ka (2.958 keV) X-ray is most likely. If that photon ESCAPEs the detector without causing ionization, then your signal will contain less ionized electrons. The process is such that during the ionizing of multiple Argon atoms, the photon will excite one Argon atom