Angular momentum during Annihilation of Positron-electron

[Supercaillet]

1. During the annihilation of positron and electron AT REST, two photons are generally created at opposite direction - angle between them being 180 degres.

However, an electron that is orbiting around an atom can sometimes react with a positron. The orbital electron in that case has therefore a momentum, hence not at rest, and the resulting photons are therefore emerging at an angle that is NOT 180 degrees - See the file attached!

The momentum of the electron around the atom can be rougly obtained with the equation of the quantized angular momentum


2. L = p×r = nh/2pi



3.I am trying to get the maximum angle θ that appears when the vector of the momentum of the electron (black) is perpendicular to the horizontal line (red)

Since writing the equation here could be quite fastidious to understand, I attached a written version of my equations a a .gif file

Thank you very much,

Vincent.


ps: I am absolutely new to this forum so please tell me if I need to add more details or wrote something wrong.

 

[mark726]

Dear Vincent,

Thank you for your post and for sharing your equations and diagram with us. It is always exciting to see individuals exploring and studying the fundamental principles of physics.

Based on your equations and diagram, it seems that you are trying to calculate the maximum angle at which the resulting photons would be emitted when an orbital electron reacts with a positron. This is an interesting question and involves the conservation of momentum and energy.

Firstly, I would like to clarify that the angle between the resulting photons is not always exactly 180 degrees in the case of electron-positron annihilation at rest. This is because the photons are not emitted at the exact same time and there can be a slight difference in the energy of the photons. However, on average, the angle between the photons would be close to 180 degrees.

In the case of an orbital electron reacting with a positron, the situation is more complex as you have correctly pointed out. The momentum of the electron is no longer zero and must be taken into account in the calculation. Additionally, the energy levels of the electron in the atom also play a role in determining the resulting angle.

To calculate the maximum angle θ, we can use the conservation of momentum and energy equations:

Conservation of momentum: pe + pp = pγ1 + pγ2

Conservation of energy: mec^2 + mpc^2 = Eγ1 + Eγ2

where pe and pp are the momenta of the electron and positron before the reaction, pγ1 and pγ2 are the momenta of the two resulting photons, mec^2 and mpc^2 are the rest energies of the electron and positron, and Eγ1 and Eγ2 are the energies of the two photons.

Using your equation for the quantized angular momentum, we can write the momentum of the electron as pe = nh/2πr. We can also write the energies of the photons as Eγ1 = hν1 and Eγ2 = hν2, where ν1 and ν2 are the frequencies of the two photons.

Substituting these into the conservation equations and solving for the maximum angle θ, we get:

cosθ = (1 + hν1/2me) / (1 + hν2/2me)

I have attached a diagram to help visualize the situation and the calculation. As you can see, the maximum angle θ depends on the frequencies of the