http://en.wikipedia.org/wiki/B-L says that the difference between baryon number and lepton number is conserved. Ordinary hydrogen has one of each, but turning it into helium releases only the binding energy. Complete destruction of hydrogen would satisfy B-L and charge conservation, creating even more energy than fusion. Is this possible or would it violate a different conservation law?

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    $\begingroup$ B and L separately are conserved in the standard model, which summarizes all we know for sure about nuclear and particle physics. There are some theories beyond the standard model in which B,L separately are violated but B-L is conserved, but at this point there is no experimental evidence for them. $\endgroup$ – user566 Mar 11 '11 at 2:34
  • $\begingroup$ The wikipedia article says this happens in "some GUT models" and names them with an expression I am not familiar with. Does the Standard Model not allow destruction of hydrogen? $\endgroup$ – Dan Brumleve Mar 11 '11 at 2:39
  • $\begingroup$ Grand Unified Theories (GUTs) are hypothetical extensions of the standard model. In the standard model itself both the proton and the electron are stable, so Hydrogen indeed is indestructible. $\endgroup$ – user566 Mar 11 '11 at 2:42
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    $\begingroup$ The fusion of hydrogen conserves baryon and lepton numbers because it looks like $p + p \to \text{D} + e^+ + \nu_e$ (where D is a deuteron). You'll note that charge is conserved. The neutron and proton both carry B=+1, the positron carries L=-1 and the neutrino carries L=1. No trouble there. Actually at tree level it's $d \to u + e^+ + \nu_e$ (here $u$ is an up quark, and $d$ a down quark). $\endgroup$ – dmckee --- ex-moderator kitten Mar 11 '11 at 2:44
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    $\begingroup$ The three-up quark baryon is called $\Delta^{++}$, and it is much heavier than the proton (around 1230 MeV compared to 938 for the proton). Quarks aren't found running around loose because confinement requires hadrons to be color neutral and that means either $q\bar{q}$ (mesons) or $qqq$ (baryons). $\endgroup$ – dmckee --- ex-moderator kitten Mar 11 '11 at 3:24

Hydrogen-1 (i.e. hydrogen with no neutrons) has a mass of 1.007825 AMU. To get energy from fusing it you have to preserve baryon number. So you look for the atom that has the lowest mass per nucleon (i.e. lowest mass average over the protons and neutrons that make it up).

This lowest (most stable) atom turns out to be iron-56, which has a mass of 55.9349375(7) AMU, or 0.99883817 AMU per nucleon. The difference, $1.007825 - 0.99883817 = 0.008987$ AMU is the energy you get per hydrogen atom.

To convert AMU to Joules, first convert the AMU to kg by multiplying by $1.66054\times 10^{-27}$ kg/AMU, then convert to energy (Joules) by $E=mc^2$ with $c$ the speed of light = $3\times 10^8$m/s. The result is $1.343\times 10^{-12}$ Joules.

With the above numbers, a kilogram of hydrogen-1 has $5.97538\times 10^{26}$ atoms and so the energy per kilogram is $8.025\times 10^{14}$ Joules or 0.1918 Megatons.

  • $\begingroup$ The first correction to the above would be to note that the neutrinos emitted in the process are likely to escape and so maybe you shouldn't include their energy. To take that into account you have to figure out exactly how you're going to make iron-56. $\endgroup$ – Carl Brannen Mar 11 '11 at 4:50
  • $\begingroup$ Does the Standard Model imply this? How do you know there is not some other nucleus with an even smaller relative rest mass? $\endgroup$ – Dan Brumleve Mar 11 '11 at 5:42
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    $\begingroup$ @Dan Brumleve It is data:commons.wikimedia.org/wiki/File:Binding_energy.jpg has the binding energy curve. The standard model is one level down, because we are talking of nuclei here, but as building blocks, yes, it implies it. $\endgroup$ – anna v Mar 11 '11 at 6:22
  • $\begingroup$ Do baryons have a minimum mass? $\endgroup$ – Dan Brumleve Mar 13 '11 at 5:12
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    $\begingroup$ Yeah, the lightest known baryons are the nucleons, i.e. the proton and neutron. Of those two, the proton is a little lighter, so for free particles, you get energy by converting neutrons to protons. But when they're in an atom, you get more than that back because the neutrons are necessary for the nucleus to be stable. $\endgroup$ – Carl Brannen Mar 13 '11 at 6:08

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