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Particle Pair Annihilation

  • #1
1. Problem Statement What is the wavelength of each of the two photons produced when a particle pair of a positron and electron accelerated at 25GeV collide? Can the net result of this annihilation be only ONE single photon? Why or why Not?



Homework Equations


Etotal(initial) = Etotal(final)
Ptotal(initial)=Ptotal(final)
Qtotal(initial)=Qtotal(final)


3. My attempt to the solution has been based on the three conservation laws which I mathematically described above, i only need a confirmation if correct.

Etotal(initial) = E(e+) + E(e-)= [mc^2 + K+] + [mc^2 + K-] = 50GeV + 2mc^2 =50 GeV + 1.022 MeV =5.0001022 × 10^10 eV

Etotal(final) = hf_1 +h_f_2 = 2hf ==> f=6.0451*10^24 Hz; lambda=c/f =4.959*10^-17 meters

Each photon will have a wavelength of 4.959*10^-17 meters.


One photon cannot result from this particle annihilation because in this case momentum would not be conserved. Since momentum is a vector quantity and after the collision all that remains is energy which is scalar there is no reasonable way that the P_initial vector be conserved. When there are two or more photons the net resultant vector of all momenta from each photon must add to zero. (is this explanation plausiable ... is there any math that can be shown???) What happens if the net P_initial is zero??? In this case the total initial momentum is probably zero since both particles are accelerated with the same energy and in opposite directions their velocity vectors sum adds to zero. Any thoughts???
 
Last edited:

Answers and Replies

  • #2
318
0
Are the last questions yours or are they part of the problem?

You could try using 4-vector (if you know them).
 
  • #3
No. They're my questions, the answers of which I think are an attempt to better answer the "Why or Why Not" part of the problem.

It makes sense that the initial momentum before collision would be zero:
p_initial = mv/(1-v^2/c^2) [tex]\hat{x}[/tex] + mv/(1-v^2/c^2) [tex]\hat{-x}[/tex] = 0

p_final of the photon may be derived by Einstein's energy formula:

E = [(pc)^2 + (mc^2)^2]^0.5. Without any mass ONE photon would have momentum given by:
p = E/c which is non-zero therefore contradictory to p_initial and momentum would not be conserved if one photon were emitted.

This in turn raises another question: what is the direction of this momentum and why should two or more emitted photons conserve momentum while ONE would not?

P.S. Sorry. I don't know the 4-vector.
 
  • #4
318
0
Your answer is correct and your reasoning is too.

You have to respect both energy and momentum conservation. Usually you go into a coordinate system that is easy to calculate, i.e. where the total initial momentum is zero. Then the final two photons will have momenta equal in magnitude but opposite in direction. One photon cannot satisfy this relation.
 

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