Widdekind Posted September 24, 2012 Posted September 24, 2012 From QM, the binding energy of electrons onto protons is ("Rydberg energy"): [math]U = \frac{1}{2} \left( m_e c^2 \right) \alpha^2[/math] and the radius of those bound electrons is ("Bohr radius"): [math]R = \frac{\hbar c}{m_e c^2} \frac{1}{\alpha}[/math] So, if the EM interaction were "full strength" ([math]\alpha \rightarrow 1[/math]), as the Strong interaction between quarks in nucleons (nearly) is, i.e. "if (virtual) photons were as strong as gluons"; then (hydrogen) atoms would be ~137x smaller (10-12m), and their binding energies would be hundreds of KeV. Now, nuclear interactions are characteristically MeV in strength. So, an "un-fine-tuned" EM interaction, at "full strength", would empower atomic interactions, between electrons and their capturing nuclei, to disrupt nuclei. Electrons could fission (larger) nuclei, upon emission, of binding energy photons, during electron capture. Perhaps in some "philosophical sense", the "de-tuning" or "dialing down" of the strength of the EM interaction, reduces atomic energies to far far less than nuclear energies, energetically segregating the former from the latter, so that larger nuclei can form, and capture electrons to participate in complex chemistries, without being broken apart by the (virtual) photons involved, in those chemical reactions, occurring "way way out" in the fluffy "electron atmosphere" of electron wave-functions, surrounding the dense tight nuclear core (a little like the 10-15 Mearth "super-earth" in the core of Jupiter's extended gaseous "super-sky") ? -1
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