Average Kinetic Energy of Ions after EI in MS?

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SUMMARY

The discussion centers on the average kinetic energy imparted to ions during electron impact ionization (EI) in mass spectrometry, specifically at 70 eV. It is established that for a hydrogen atom, the remaining 55 eV of kinetic energy is not significantly transferred to the proton, which receives less than 1 eV due to its mass relative to the electrons. The principle that lighter particles tend to absorb more kinetic energy than heavier ones is highlighted, indicating that in cases involving heavier ions, the kinetic energy transfer is even less.

PREREQUISITES
  • Understanding of electron impact ionization (EI) in mass spectrometry
  • Knowledge of kinetic energy transfer principles in particle collisions
  • Familiarity with mass-to-charge ratios in ionized particles
  • Basic concepts of ion fragmentation and its effects on energy distribution
NEXT STEPS
  • Research the principles of kinetic energy transfer in particle physics
  • Explore the effects of mass-to-charge ratios on ionization efficiency
  • Study advanced mass spectrometry techniques and their applications
  • Investigate the role of ion fragmentation in mass spectrometry data interpretation
USEFUL FOR

This discussion is beneficial for mass spectrometry researchers, analytical chemists, and students studying ionization processes and energy transfer in particle physics.

BlakeLeonard
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Hello,

From what I've been reading, 70eV electrons are the standard for electron impact ionization in mass spectrometry. I'm trying to get a feel for how much kinetic energy on the average is imparted to an ion. I know this gets complex, when you start talking about multiple fragments, but in just the simplest case, a Hydrogen atom, how much of the remaining 55eV is converted to kinetic energy of the ion? Thanks.

Blake
 
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Well, I don't know what fraction of the KE of the electron is imparted to the H-ion (proton), but I can tell you that, at only 56 eV available for KE, the proton will not even "feel" it. One complication is that there are two electrons and one proton in the final state (which you alluded to). One consideration is that the smaller particle tends to take the larger portion of the KE. For instance, when M>>m, then the lighter projectile tends to take almost all of the available KE, and the heavier target tends to obtain only a fraction on the order of m/M. In the case of a hydron atom, that is a fraction of only 1/2000. So, I would expect that the proton doesn't even obtain a single eV. For the case of a heavier ion, this is even much smaller.
 

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