# What has $E = mc^2$ to do with nuclear powerplants?

In life, when you talk about nuclear energy, there always happens to be a guy who says that famous Einstein's equation. "Yeah, they just convert mass to energy, $E = mc^2$ ya know?"

When I think about that, all I learned about nuclear power resembles dominoes arrangements. You tonk a block and it falls. On its way, it tonks other dominoes and when it falls it releases energy (sound waves).
Quite same in the nuclear physics. You send slow neutron to a core. The core absorbs it, breaks and sends another neutrons and energy (electromagnetic waves).

So in the end, I see no domino blocks disappearing in this game. All we do is, that we tonk domino arrangements that has been built by old stars long time ago.

So why is this equation related to nuclear power? What mass disappears in nuclear power plants?

-
Dear @Tomáši, please don't get offended but what's amusing is that your discussion of the fate of energy is completely wrong even in the case of the ordinary dominos. Dominos wouldn't just fall and create lots of noise and fun "for free". The energy needed to achieve such a thing has to come from somewhere - it comes from the gravitational potential energy of the domino pieces that are standing at the beginning but lying at the end. Similarly, the gravitational potential energy is replaced by the nuclear binding energy - also equal to $E=\Delta mc^2$ calculable from the mass difference - here. – Luboš Motl May 29 '13 at 9:02
And otherwise no one is talking about disappearing protons, neutrons, or dominos here. The point is that the total mass of the nuclei changes - the average mass of the proton or nucleon in these nuclei - is changing by something of order 0.1% (fission) to 1% (fusion) when nuclear reactions take place. In fact, even in the case of the dominos, the total mass of the domino pieces after they fall is a tiny bit smaller than the mass at the beginning, by $\Delta m = E/c^2$ where $E$ is the gravitational potential energy that was released (e.g. via sound). – Luboš Motl May 29 '13 at 9:04
I did my best not to get offended. But still I don't think you should consider funny, that I don't know the answer to my question - I'm doing my bes to ask questions in a smart and useful way. But I can't predict the answers to ask questions with no mistakes in them. – Tomáš Zato May 30 '13 at 1:38
– Qmechanic Jun 11 '13 at 0:20

However, the amount of mass lost is very small, even compared to the masses of the constituent particles. A good overview is given in the Wikipedia article on mass excess. Basically, the mass of a nucleus will in general be a little bit off from the sum of the masses of the protons and neutrons inside it. This is because there is a binding energy holding the nucleus together, and your standard $E = mc^2$ gives the equivalent mass for this energy.
In the fission of uranium-235, $${}^{235}_{\phantom{0}92}\mathrm{U} + {}^1_0\mathrm{n} \to {}^{236}_{\phantom{0}92}\mathrm{U} \to {}^{141}_{\phantom{0}56}\mathrm{Ba} + {}^{92}_{36}\mathrm{Kr} + 3\ {}^1_0\mathrm{n},$$ the total rest mass of the products is slightly less than of the reactants. This is true even though there are the same number of protons (92) and neutrons (144) before and after. So it is not as though an entire nucleus disappears, or even an entire proton or neutron. The lost mass comes from the binding energy.