Average Kinetic Energy of Ions after EI in MS?

AI Thread Summary
In mass spectrometry, 70 eV electrons are standard for electron impact ionization, but the average kinetic energy imparted to ions can vary. For a hydrogen atom, only a small fraction of the remaining 55 eV is converted to kinetic energy, with the proton likely receiving less than 1 eV due to its mass compared to the electrons. The distribution of kinetic energy favors the lighter particle, meaning the proton, as a heavier target, obtains minimal energy. This principle applies more significantly to heavier ions, resulting in even lower kinetic energy transfer. Understanding these dynamics is crucial for interpreting ion behavior post-ionization.
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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