 #1
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Homework Statement:
 1. Proton of energy ##76,4\mbox{GeV}## colides with the proton with the proton that staying in the rest. How much proton  antiproton pairs is possible to generate?
Relevant Equations:

[tex]E^2=E_0^2+p^2c^2[/tex]
##E_0=0.94\mbox{GeV}##  rest energy of proton and antiproton
Momentum ##\vec{p}## before collision is momentum of proton of the energy ##E=76.4\mbox{GeV}##. Law of conservation of energy is
[tex]E+mc^2=E_1+E_2+...+E_n[/tex]
[tex]mc^2=0.94\mbox{GeV}[/tex]
We could generate only even number of particles after collision because of law of conservation of electric charge. Also there need to be two more protons then antiprotons. So, ##n## is even number and there is two more protons then antiprotons after the collision. Conservation of momentum gives
[tex]\vec{p}=\vec{p}_1+\vec{p}_2+\vec{p}_3+...+\vec{p}_n[/tex]
I am not sure what to do next?
Also because
[tex]\frac{E+mc^2}{mc^2}=82,28[/tex]
we now then number of particles is smaller then ##82## because law of conservation of momentum.
[tex]E+mc^2=E_1+E_2+...+E_n[/tex]
[tex]mc^2=0.94\mbox{GeV}[/tex]
We could generate only even number of particles after collision because of law of conservation of electric charge. Also there need to be two more protons then antiprotons. So, ##n## is even number and there is two more protons then antiprotons after the collision. Conservation of momentum gives
[tex]\vec{p}=\vec{p}_1+\vec{p}_2+\vec{p}_3+...+\vec{p}_n[/tex]
I am not sure what to do next?
Also because
[tex]\frac{E+mc^2}{mc^2}=82,28[/tex]
we now then number of particles is smaller then ##82## because law of conservation of momentum.
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