Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

Suppose there is a material with heavy nuclei attached on its surface, presumably binded by the outer shell electrons. Now, the surface is exposed to a cold positron gas, which annihilates against some of these outer shell electrons.

What might happen with the heavy nuclei? do they have to become free from the material surface? can they become binded by inner shell electrons?

Also, i presume that as the positron gas annihilates the outer shell electrons, and further assuming that the heavy nuclei remain attached somehow to the surface, these heavy nuclei will increase their effective charge, so there might be an equilibrium condition where the repulsion of the heavy nuclei wards off the positron gas from further annihilating against the inner core electrons. Can this be validated heuristically? Is it just a plain wrong assumption?

share|improve this question

2 Answers 2

As the nuclei will become more positive they will bind better with the outer electrons of the surface they have been attached to previously, just by electrostatic forces.

It is improbable that your cold positrons will be able energetically to scatter on a second electron and annihilate it. The process can be calculated. It has been for positron Helium scattering>

share|improve this answer
    
thanks Anna, this is interesting. Do you think the electrostatic repulsion of the heavy ions in the surface and the positron gas could work to keep the positrons contained and stable up to significant densities (i.e: higher than what is attainable with a Penning trap)? –  user56771 Nov 6 '12 at 14:17
    
Ι cannot tell . Certainly the positive charge from the ionized nuclei will contribute to the electric field but it needs an expert in penning traps to give an answer. –  anna v Nov 6 '12 at 16:08

As the positron gas density grows, it does not stay cold, since you are doing work against it to keep it compressed. This energy spreads evenly over the degrees of freedom in the gas, so you'll get some highly energetic positrons to overcome the electrostatic potential of the heavy nuclei. It also depends of course on how thick the heavy nuclei layer in the inside of the container is.

It would still be interesting to know quantitatively at what density/pressure does this method stops being efficient. If i come up with a rough estimate, i'll post it later

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.