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I am only 14, so I don't know much about physics, and I would appreciate help on this topic.

A photon, is said to have 0 mass. However, they carry energy, as electromagnetic radiation. Albert Einstein proved that $E = mc^2$. But if you plug the data in, $E = 0\times (299791)^2$, you try to multiply the speed of light squared by $0$. Therefore, the energy of a photon would be $0$. How is that possible? Since all the photon does is carry energy?

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Hi Ze Photon. Welcome to Physics.SE. I think the content for an explanation to your question is too much for a 14-year old. If you still can't resist, here are a few questions already here: physics.stackexchange.com/q/2229 and physics.stackexchange.com/q/3541 –  Waffle's Crazy Peanut Feb 7 '13 at 16:02
    
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$E=mc^2$ is only for things which don't have velocity. $E^2=m^2c^4+p^2c^2$ is correct. But when the velocity is $0$ ($v=0$) then the momentum is $0$ ($p=mv=0$) so you get the equation $E=mc^2$. –  raindrop Jul 13 '13 at 6:59
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marked as duplicate by dmckee Feb 7 '13 at 16:11

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up vote 5 down vote accepted

The relation $E=mc^2$ only works for particles at rest, which is evidently not the case for photons. In the general case, the relation is $$E^2=m^2c^4+p^2c^2$$ for a particle with momentum $p$. (Note, though that the momentum is not necessarily $p=mv$ as in the newtonian case! See for instance If photons have no mass, how can they have momentum?)

For a massless particle, then, $E=pc$.

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