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Density and muon

  1. Jan 28, 2013 #1
    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
  2. jcsd
  3. Jan 28, 2013 #2


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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.
  4. Jan 28, 2013 #3
    Thanks! I completely missed the quantization of angular momentum. When r~m-1 then E~m*r-2~m3.
  5. Jan 28, 2013 #4


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    Just keep in mind that it's not just angular momentum. L=0 orbitals shrink by the same factor.
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