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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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