Difference between revisions of "ArCO2 IonizationPhysics"

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(New page: <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)
 
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<math>\bar{E}=</math> 27 eV = average energy to ionize and electron in an Argon Atom
 
<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
 
<math>\bar{N_{PE}} = \frac{E_{\gamma}}{27 eV} =</math> average number of photoelectrons produced
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Quenching Gas:
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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
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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.
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If the quech gas is CH4 then
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CH4+ <math>\Rightarrow</math> H2 + CH2
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;Argon Escape peak:
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: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.

Revision as of 02:51, 19 December 2008

[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.