How Do Gamma Rays Result from Electron-Positron Annihilation?

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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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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.
 
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).
 
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) ??
 
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?
 
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
 
Yes so it is already conserved. So you are missing another conservation law in order to have two equations and two variables.
 
Energy conservation:

ϒmc2=2P c cos θ

is it correct?
 
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.
 
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