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The available energy ea when the proton and the antiproton have equal energies of 450gev is 29 GeV.
How to find the available energy after the collision?
The available energy can be found out with the following formula.
[tex]E_a=\sqrt{2E_tE_k++(m_1c^2)^2+(m_2c^2)^2}[/tex]
Here C is the speed of light (3×10⁸ m/s), Et is the energy of proton, Ek is the rest energy.
A proton and an antiproton have equal energies of 450gev. The particles collide head-on. The rest energy of the proton is 938mev
Thus, put the given values in the above formula,
[tex]E_a=\sqrt{2(450\times\dfrac{10^9eV}{1MeV})(938\times\dfrac{10^6eV}{1MeV})+(938\times\dfrac{10^6eV}{1MeV})^2+(938\times\dfrac{10^6eV}{1MeV})}\\E_a=29\rm\; Gev[/tex]
Thus, the available energy ea when the proton and the antiproton have equal energies of 450gev is 29 GeV.
Learn more about the proton here;
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