How much energy is released from an electron-positron annihilation?

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The energy released from electron-positron annihilation is primarily the mass energy of the particles, totaling 1.022 MeV, along with any kinetic energy they possess. While the electromagnetic attraction between the two particles is significant, it does not lead to infinite energy release, as suggested by Coulomb's Law. The annihilation occurs not necessarily when the particles "touch," but rather at a distance determined by electromagnetic interactions. Quantum Electrodynamics (QED) can describe this process, but basic conservation of energy principles suffice for understanding the annihilation phenomenon. Overall, the energy released is a combination of mass-energy and kinetic energy, influenced by their separation distance.
thomas9459
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More specifically, I am wondering what the energy released is when the electromagnetic force is also included in this calculation. In the case where the electron and positron are at a distance d, the energy released would be the mass energy (1.022 MeV) plus the energy that resulted from the electromagnetic attraction. Using Coulomb's Law, the energy released should be infinite, which is obviously not the case. What does QED say about the mater? Has an experiment which measured the energy released accurately enough to see the extra energy from the electromagnetic attraction?
 
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That still doesn't answer the question of how much energy is released from the annihilation.
 
thomas9459 said:
That still doesn't answer the question of how much energy is released from the annihilation.

I'm a layman but I'd think that since the both particles are being annihilated, that no energy is released; don't they just cancel each other out?
 
CyberShot said:
I'm a layman but I'd think that since the both particles are being annihilated, that no energy is released; don't they just cancel each other out?
That violates conservation of energy and is certainly not true at all.

thomas9459 said:
What does QED say about the mater? Has an experiment which measured the energy released accurately enough to see the extra energy from the electromagnetic attraction?
Pal, you don't have to drag in QED in order to describe this phenomenon! This is perfectly describable using basic physics - the concept of conservation of energy (or rather, mass-energy in this case).

Imagine I have a system consisting of an electron and positron at a distance d from each other. The total energy of the system at that point is
m_{e}c^{2} + m_{p}c^{2} + KE_{e} + KE_{p} - k \frac{e^{2}}{d^{2}}
By conservation of mass-energy, the total energy released when they annihilate is the same.
 
Fightfish said:
m_{e}c^{2} + m_{p}c^{2} + KE_{e} + KE_{p} - k \frac{e^{2}}{d^{2}}

So a better question is at what distance do they annihilate. My impression was it happens when the electron and positron "touch", but since they are point particles, this would happen at d=0, which results in a division by zero.
 
you try to mix quantum mechanics with classical mechanics in not a good way and thus you result in this problem.
The phenomenon of annihilation is an electromagnetic one, thus you truly have electromagnetism already inside.
Generally it doesn't happen when they "touch". It happens in whatever a distance they are according to the possibilities of electromagnetism interaction (thus the coupling constant). Thus you have annihilation as you also have the pair creation of a photon of E_g ~> 1MeV (eg Na22)
 
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