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# Does "E=mc2" theory rules out in the case of "LARGE HYDROGEN COLLIDER" process?

when the particles gets collided in "LHC"
Energy released during that collision does equate with "E=mc2" theory (or) "E=mc2" rules out!!!
Shashank Moghe • Mar 31, 2016
when particles in "LHC" collide a lot of energy(E) is released.Now in E=mc2 on substituting values for E,m,c does it satisfies or rules out !!!
Shashank Moghe • Apr 4, 2016
E=mc2 gives the maximum energy potential of a rest mass "m". This is the limit of the energy you can get from this mass. When particles collide and energy is released, a fraction of the mass is converted into energy. Similar conversion occurs in many chemical and nuclear reactions. If you are asking whether Conservation of Energy is violated, it is not.
thats the real generalization
but how does collision of a small or negligible mass of those particles would equate to enormous energy(E) released in E=mc2
Shashank Moghe • Apr 5, 2016
c = 299792458 m/s (source: Google)
c^2 = 89875517873681764 m^2/s^2

Multiply by what you might call a "negligible" mass with this, lets say a milligram (10^-6 kg).

E = 10^-6 * c^2 = 89875517873.681764 Joules

Not small, is it?
the calculation was perfect!
but the mass of particle means the mass of proton(mp)
mp=1.6726219*10^-27 kg that matters.

now the calculated value moves on to
E=[1.6726219*10^-27]*[89875517873681764]