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?
Answer
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.
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