If the proton and antiproton come closer to each other how much energy released
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It just for calculation but it not experimental proof
The energy released = total energy of proton and anti-proton.
Rest energy of a proton/anti-proton = 938.257 MeV (Mega-electronvolts)
now E = mc^2
2 * 938.257 = 1876.514 MeV ( cause there are 2 particles)
1876.514 * 10^6 = 1876514000 eV (get from MeV to eV
1 eV = 1.60 * 10^-19J (convert to joules using charge of an electron (because we're in electron volts))
So 1876514000 eV = (1876514000 * (1.60 * 10^-19)) J
= 3.00 * 10^-10 J
The energy released = total energy of proton and anti-proton.
Rest energy of a proton/anti-proton = 938.257 MeV (Mega-electronvolts)
now E = mc^2
2 * 938.257 = 1876.514 MeV ( cause there are 2 particles)
1876.514 * 10^6 = 1876514000 eV (get from MeV to eV
1 eV = 1.60 * 10^-19J (convert to joules using charge of an electron (because we're in electron volts))
So 1876514000 eV = (1876514000 * (1.60 * 10^-19)) J
= 3.00 * 10^-10 J
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