0
$\begingroup$

Lets say two hydrogen fuse together, where does the energy released go?

Is it carried away as momentum imparted on the helium atom? Is it carried away in neutrinos? Is it carried away as gamma rays? Or is it a combination of all of the above in specific proportions?

Also, if neutrinos are massless, can they carry energy away as momentum, or is that not possible?

$\endgroup$
3
$\begingroup$

Is it carried away as momentum imparted on the [product] atom? Is it carried away in neutrinos? Is it carried away as gamma rays?

All of these can happen, and in general nuclear reactions will output their energy via a combination of these. The specific combination, of course, depends on the specific reaction.

Also, if neutrinos are massless, can they carry energy away as momentum, or is that not possible?

As it happens, neutrinos are not massless (or at least some of them are not; the jury is still out on that but the situation is slightly more complicated). For the purposes of most nuclear reactions, this mass is small enough that you can neglect it, but this does not pose a problem. In general the energy of a particle with mass $m\geq0$ and momentum $p$ is equal to $$E=\sqrt{m^2c^4+p^2c^2}.$$ If $m=0$, or it is very small, then the particle can still carry momentum and an associated energy. In general, neutrinos can carry away a significant amount of energy via this mechanism.

$\endgroup$
  • $\begingroup$ Thanks. Am I right in thinking that you are saying that the amount of energy carried away by neutrinos is negligible? $\endgroup$ – user2800708 Sep 30 '15 at 15:02
  • $\begingroup$ @user2800708 Not if it is moving very fast. Particles like those in the LHC carry a lot of energy though I'm not sure about nuclear reactors. $\endgroup$ – Horus Sep 30 '15 at 15:26
  • 1
    $\begingroup$ @user2800708 "Am I right in thinking that you are saying that the amount of energy carried away by neutrinos is negligible?" Generally, no. In the case of beta decays they can carry as much as a third of the energy. I don't have any figures for fusion events on tap, but you can expect them to be significant. Indeed the energy gains for stellar fusions cycles are always quoted not with the mean neutrino contribution already taken out. $\endgroup$ – dmckee Sep 30 '15 at 15:38
  • $\begingroup$ @dmckee That last bit is slightly confusing. Stellar fusion energy gains are quoted without neutrino contributions, then? $\endgroup$ – Emilio Pisanty Sep 30 '15 at 15:43
  • $\begingroup$ @EmilioPisanty Yeah. Your version is better. I wasn't sure that I should write about that though. Because I suspect there are corners fo the field where the practice differs. I know that during super novae a fraction of the neutrino energy gets deposited, and I suspect that really massive stars probably collect a measurable fraction of it too. $\endgroup$ – dmckee Sep 30 '15 at 15:46
-3
$\begingroup$

It is not carried away at all. It is now stored in the nuclear bond.

$\endgroup$
  • 1
    $\begingroup$ No, the energy in the bond after the reaction is less (essentially more negative) than the energy before. Otherwise, fusion and fission (as well as chemical reactions) would never be exothermic. $\endgroup$ – Bill N Sep 30 '15 at 16:23
  • $\begingroup$ Whether fusion is exothermic (like in the case of stars) is really the question. $\endgroup$ – TheDoctor Sep 30 '15 at 16:27
  • 2
    $\begingroup$ "Whether fusion is exothermic (like in the case of stars) is really the question." Er...the answer is well known for a large number of reactions and for the case the OP proposes it is know to be highly exothermic. $\endgroup$ – dmckee Sep 30 '15 at 17:52
  • $\begingroup$ I think the "binding energy" is negative. $\endgroup$ – user2800708 Oct 1 '15 at 8:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.