In a hydrogen atom, when a muon is substituted for the electron, its orbit is much nearer to the proton (with theoretical advantages for ‘cold’ or ‘room-temperature’ fusion evidently).
Why does the muon in the muon-substituted hydrogen atom orbit so much closer to the proton? I assume it’s because the muon is more massive, but how does the increased mass translate to decreased orbital distance?
Is it due to the Uncertainty Principle?, i.e. the greater muon mass means it has more momentum and thus less uncertainty in its position than the native atom’s electron. The increased precision in the muon’s position means it can occupy, and does occupy, a smaller volume compared to an electron. Is that sort of the mechanism?
Consider a classical circular gravitationally-bound orbit of a particle moving at fixed speed. The force provided by gravity is proportional to m/r[sup]2[/sup] and the force required to maintain circular orbit is proportional to m/r. This means that the radius of this orbit is proportional to a constant, which means in particular that it doesn’t depend on mass.
Now consider a Coulombic orbit. The strength of the Coulomb force is proportional to q/r[sup]2[/sup], but the force required is again proportional to m/r. Here, the m’s don’t cancel, and the radius of a Columbic circular orbit is proportional to 1/m.
Voilà? Yes, if you aren’t being too picky.
Otherwise, you have to deal with the speed dependence of the radii, which I didn’t write in explicitly above (since I assumed some given, fixed speed). In the classical case, I can give my orbiting object any speed I please, so any radius is possible. In the quantum case, the energies are quantized, meaning that the (effective) speeds are quantized, too. The proper way to deal with this is to solve Schrödinger’s equation with a Coulomb potential to find the energy levels of the system and to figure out what the semi-classical speed would be of an orbiting particle in, say, the lowest energy level. A not-too-rigorous shortcut around this would be just to say: hey, it’s a quantum thing, so it needs to involve h-bar, which has units of angular momentum; so let’s say that the angular momentum mvr of our object is quantized in units of h-bar. Thus, mvr=n(h-bar). Choosing n=1, because why not, gives an expression for v that can be substituted into the fuller expression for the radius (which is r proportional to 1/(mv[sup]2[/sup]) ). The result after this substitution is that r is proportional to (h-bar)[sup]2[/sup]/m. So, it still has the 1/m behavior even in the quantum realm.