Electron-positron annihilation

  • Thread starter Guido Diforti
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In summary, in a collision between an electron with kinetic energy K=1Mev and a positron at rest, both particles mutually annihilate and create two gamma rays with equal energy. The energy E, linear momentum P, and emission angle θ of the gamma rays can be calculated using the equations K=γmc^2-mc^2, ϒmc^2=2Pc, and ϒmc^2+mc^2=2Pc. The conservation of energy and linear momentum must be taken into account when solving for these variables.
  • #1
Guido Diforti
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I need help with a problem and I would appreciate if you help me with it...

An electron with kinetic energy K=1Mev collides with a positron at rest. In the collision, both particles mutually annihiliate, creating two gamma rays with equal energy, each of them traveling in equal angles θ in the direction of the movement of the electron. Find the energy E, the linear momentum P and the emmision angle θ of the gamma rays.

Thanks
 
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  • #2
Hello Guido, welcome to PF :smile: !

If this is homework, you 'd better post it in the homework part of the forum. They also have a nifty template, giving you the opportunity to quote the relevant equations and present an attempt at solution, so you can be given adequate asssistance. much more useful than giving you all the answers on a plate -- which seems nice, but isn't really helping you at all.
 
  • #3
Hi Guido, welcome to PF!

I have moved your question to the homework section, however before we can answer we need you to show your attempt (even if you know it is wrong).
 
  • #4
Thanks for moving my post.

Here's my attempt:

With the kinetic energy K i was able to calculate the velocity v of the electron:

K=ϒmc2-mc2

From the equation above : v= 0,9893c

I'm having trouble with the conservation of the linear momentum P, which I guess I must do the conservation in X and Y.

X) mvϒ= 2E/c cosθ ??
Y) ??
 
  • #5
Guido Diforti said:
Y) ??
What is the momentum in the y-direction before the annihilation? What is it after?

Are you missing a conservation law?
 
  • #6
Momentum in the y direction is 0 before and after since the movement is in the x direction :

γmvy=0=E/c sin θ + E/c sin(-θ)=0
 
  • #7
Yes so it is already conserved. So you are missing another conservation law in order to have two equations and two variables.
 
  • #8
Energy conservation:

ϒmc2=2P c cos θ

is it correct?
 
  • #9
No, there are two electrons before the annihilation. Both of them have energies.

Also, do the photon energies depend on the angle?
 
  • #10
ok so one is at rest and has an energy associated:

ϒmc2+mc2=2pc
 
  • #11
Yes, so now you have two variables and two equations. Just go ahead and solve the system.
 
  • #12
Thank you, you've been very helpful! Sorry about the mistakes, I'm new to special relativity.
 

Related to Electron-positron annihilation

1. What is electron-positron annihilation?

Electron-positron annihilation is a process in which an electron and a positron (its antiparticle) collide and convert their mass into energy, producing two high-energy photons.

2. What is the significance of electron-positron annihilation?

Electron-positron annihilation plays a crucial role in particle physics, as it helps scientists understand the fundamental properties of matter and antimatter. It also has practical applications in medical imaging and cancer treatment.

3. What conditions are necessary for electron-positron annihilation to occur?

Electron-positron annihilation can only occur when the particles have enough kinetic energy to overcome their mutual electric repulsion and come into contact with each other.

4. How is electron-positron annihilation detected?

Electron-positron annihilation can be detected through the presence of high-energy photons that are produced during the process. These photons can be detected using specialized instruments such as gamma-ray detectors.

5. Can electron-positron annihilation be reversed?

No, electron-positron annihilation is an irreversible process. The energy produced from the annihilation cannot be converted back into matter.

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