Why is a pair of an electron and a positron unstable?

[Mayan Fung]

We know that an electron and a positron will annihilate and emit gamma ray. But the electron and positron possesses initial kinetic energy meaning that it is difficult for them to really collide in each other. Just like Earth is not dropping into the Sun even with the gravitational pull. So I am curious how scientists can annihilate the two. And what is the scale of their distance such that annihilation occurs?

 

[Drakkith]

But the electron and positron possesses initial kinetic energy meaning that it is difficult for them to really collide in each other. Just like Earth is not dropping into the Sun even with the gravitational pull.

Their attraction results in an acceleration of both charges, which causes the emission of electromagnetic radiation. This robs energy from both of them and causes them to very quickly spiral into each other.

As for the distance scale where annihilation occurs, I admit I don't know.

 

[Mayan Fung]

Their attraction results in an acceleration of both charges, which causes the emission of electromagnetic radiation. This robs energy from both of them and causes them to very quickly spiral into each other.

As for the distance scale where annihilation occurs, I admit I don't know.

We know that electrons are not emitting EM waves while "orbiting" around a proton, as long as they stay at the steady state (the orbital). Is it possible that an electron-positron system also has such a steady state so that they don't annihilate?

 

[Drakkith]

We know that electrons are not emitting EM waves while "orbiting" around a proton, as long as they stay at the steady state (the orbital). Is it possible that an electron-positron system also has such a steady state so that they don't annihilate?

To my knowledge, there is not, but I don't know the details of why. Perhaps someone else here can provide a more in-depth explanation.

 

[snorkack]

We know that electrons are not emitting EM waves while "orbiting" around a proton, as long as they stay at the steady state (the orbital). Is it possible that an electron-positron system also has such a steady state so that they don't annihilate?

There is only one such steady state, and that is 1s.
In which electron does not orbit around the proton, but oscillates through the proton.
The density of electron at the proton, though finite, is nonzero at the cusp of 1s orbital, and smaller but still nonzero at the cusp of all higher s orbitals.
p, d and higher angular momentum orbitals possesses nodes that go through proton and thus have zero density at proton.

There are many other nuclei besides positron that can annihilate an electron. Starting with Be-7, most common K-40 etc. While annihilation of 1s electrons is most common, electrons can also be annihilated from 2s and higher orbitals, as well as molecular and crystal orbitals of s character at that nucleus. But not from p, d, etc. orbitals or molecular or crystal orbitals of p, d etc. character at that nucleus.

 

[mfb]

They can annihilate directly if they come close enough (not exactly a well-defined thing in quantum mechanics, but the classical analogy is not too bad here). The probability (the cross section) decreases with increasing energy. They can also form positronium first, decay to its ground state and then annihilate from there.

 

[snorkack]

They can annihilate directly if they come close enough (not exactly a well-defined thing in quantum mechanics, but the classical analogy is not too bad here). The probability (the cross section) decreases with increasing energy. They can also form positronium first, decay to its ground state and then annihilate from there.

Or decay to any higher s orbital and annihilate from there. Or to a molecular or crystal orbital with s character at positron.
Does an unbound orbit have a well-defined orbital angular momentum (s, p etc.)?

 

[mfb]

Or decay to any higher s orbital and annihilate from there.

Not impossible but very unlikely.

Does an unbound orbit have a well-defined orbital angular momentum (s, p etc.)?

If you measure it: Sure.

 

[snorkack]

Not impossible but very unlikely.

Not in case of 2s orbital.
The radiative decay of 2s positronium into 1s is strongly (though not completely) forbidden, just as 2s hydrogen and for the same reasons.
Whereas annnihilation from 2s state is slower than from 1s, but not drastically so.
And states that can decay to either 2s and 1s (such as 3p) have no strong reason to prefer 1s over 2s.

 

[Heikki Tuuri]

https://arxiv.org/abs/hep-ph/9911410
The lifetime of a positronium "atom" is calculated with quantum electrodynamics. The amplitude for the decay is some kind of a path integral.

Since it is not a standard particle collision experiment, but involves a bound state, the positronium, people probably use some extra assumptions in the calculations.

 

[geoelectronics]

Positrons can be generated by "Pair Production", where an energetic electromagnetic photon (E.g. Gamma Ray) containing at least 1022 keV of energy interacts with the Nuclear Force of an atom's nucleus. It disappears and is replaced by an electron-positron pair which share all the energy once contained in the photon (one form of energy to matter conversion). How far the positron travels depends on the total energy it contains. At rest it will contain 511 keV of mass energy, anything above that level must be lost to the environment before annihilation can occur.
Upon annihilation the positron disappears and is replaced by two 511 keV photons, traveling in opposite directions. These photons are called neither Gamma Rays nor X-ray, but "annihilation radiation".

The second common source of positrons is radioactive beta+ decay of an atomic nucleus by the process called "positron emission" where the nucleus ejects a positron from a proton, changing it into a neutron, therefore transmuting to the element immediately lower on the Periodic Table.

F-18 is used in medical exams for its 633 keV positron emission, in a machine called PET-CT.

I do an interesting experiment with Na-22 using magnets that shows how to direct electrons from radioactive decay by using magnetic polarity to separate B- electrons (i.e. negatrons) from B+ positrons in air, into a detector several inches away.

George Dowell

 

[mfb]

At rest it will contain 511 keV of mass energy, anything above that level must be lost to the environment before annihilation can occur.

It doesn't have to, it is just the typical case. Annihilation can happen at higher energies. Particle colliders use this routinely.

 

[geoelectronics]

Correct, my description is of the common natural pair in thermalized condition. I myself haven't observed any natural e-p annihilation that didn't produce two 511 keV or so close that my instruments can't tell.

Your reference is to CERN's LEP etc., and there the two particles contain enough extra kinetic energy to the extent that physical particles are also produced.I have a question, during such a high energy collision, is there also a pair of 511 keV rays produced?

Thanks

George Dowell

 

[mfb]

Extra particles are possible, but two photons are still a common result. Energy is conserved, the photons have more energy in colliders. ~10 GeV in total in B-factories like SuperKEKB.

 

[geoelectronics]

SuperKEKB

link to a nice paper that reveals a lot about the 5 modes of e-p (or as I call it beta- / beta+ annihilation) annihilation:

https://pdfs.semanticscholar.org/3399/19854fb8b51e11e0c918ab6db7afbc93e811.pdf

George Dowell