Nuclear knockout reactions - why does a proton knockout a neutron?

In summary, the process of creating F-18 used in FDG involves the collision of O-18 with an accelerated proton, resulting in the proton knocking out a neutron. This reaction is not the desired one, but it is possible. The F-18 formed is briefly an ion but will quickly capture an electron from the environment. The proton beam used in the process is a source of positive charges, but the equipment is grounded to keep it neutral.
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Questions about the formation of F-18 from O-18; why a neutron is selectively knocked out, how do the charges balance out
I am studying the synthesis of FDG for a job interview. The process of creating the F-18 used in FDG involves the collision of O-18 with an accelerated proton -- the proton kicks out a neutron and everything adds up. I am just wondering why a neutron is specifically knocked out, why wouldn't another proton be kicked out instead? Is it 50/50 but the proton exchange is essentially no change at all? Thanks for any replies.

Bonus question: I might make this a separate post. But when the F-18 is formed it would be positively charged. I believe it quickly turns into an anion but still, the charges would not be balanced in the system (correct?). Would the solution have an overall electrical charge? Are electrons stripped from the environment? (In chemistry you usually can't just add a charged particle to a solution without its opposite charge. I.e. Na+Cl- )
 
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O-18 + p -> O-18 + p is a possible (and even likely) reaction, it's just not the reaction people are interested in. You can't control what reaction happens but you can bombard oxygen (or water) with enough protons and then separate out fluorine.

It's an ion briefly but it will quickly capture a random electron from the environment.
Your proton beam is a source of positive charges but grounding the equipment keeps it neutral.
 

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