Exploring the Relationship between Density and Muon in Bohr's Model

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Replacing electrons with muons in Bohr's model theoretically increases atomic density by a factor of 200³ due to the muon's greater mass. The radius of electron orbitals is inversely proportional to the electron's mass, meaning heavier particles occupy smaller orbits. This results in higher momentum for muons, leading to lower energy states at smaller radii. The relationship between mass and density is derived from the quantization of angular momentum, which applies to all orbitals, including L=0. Thus, while the behavior of electrons and muons differs, the underlying principles of Bohr's model support this density increase.
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In my (very brief) lecturer's notes there's written that ρ~mp~me3 (*). So.. when (hypothetically) replacing every electron with a muon (around 200me), could the density increase 2003 times? Where comes that (*) relation (in Bohr's model)?

Just in case: it's not a homework question
 
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Basically, yes. The radius of the electron orbital is inversely proportional with electron's mass. So if you replace it with a 200x heavier muon, then all of the atoms will get 200x smaller, increasing density by factor of 200³. Actually, a bit more, even, because of the added mass of the muon.

In Bohr's atom, this is very easy to understand. The orbit is integer number of wavelengths, which fixes possible values of momentum, while the heavier particle in a particular orbit will have higher momentum. So the ground state, the lowest possible momentum and energy, is attained at much lower orbit for a heavier particle.

Of course, that's not actually how the electron or muon behaves in an atom, but this is how Bohr derived the radius and relationship, which just happened to be correct. The principle is similar, however. Muon will have higher momentum in an atom, and therefore, its orbitals will be smaller, giving you higher density as a result.
 
Thanks! I completely missed the quantization of angular momentum. When r~m-1 then E~m*r-2~m3.
 
Just keep in mind that it's not just angular momentum. L=0 orbitals shrink by the same factor.
 

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