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These relations are based on the fact that both the position and the momentum distributions are centred around zero, which is in turn due to the symmetry of the atom. Given that, the width of the position and momentum distributions ($\Delta x$ and $\Delta p$) is of the same order as a typical position or momentum within those distributions ($r$ and $p$).


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The electron is the lightest lepton and the proton is the lightest baryon, so it's hard to see what reaction could occur without violating lepton number or baryon number. I suppose if proton decay (to a pion and positron) occurs then there could be a reaction to give a pion and two photons.


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The symmetries that you're missing are conservation of baryon number $B$ and lepton number $L$. We strongly suspect that baryon number is not an exact symmetry, because the universe appears to contain very many baryons and very few antibaryons. Actually, a better metric for the baryon asymmetry of the universe is to compare the baryon density to the density ...


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If you assume perfect efficiency, then the energy of dissociation of a liter of water is computed as follows: 1 liter of water, molar mass 18g, => 55.6 moles The energy needed is 237 kJ per mole (from your link - see under "thermodynamics"). 237 * 55.6 => 31.7 MJ of energy for a liter of water. In terms of power, this is 3.67 kW for one hour. This ...


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The way this is justified is as follows: We start with the uncertainty principle, which can be roughly stated as $$\Delta x \Delta p \geq \hbar$$ For this rough estimate, we will ignore some factors of perhaps $2$ or $\pi$, but we're interested in some order of magnitude, not the exact result. Now, our second assumption will be that the ground state of the ...



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