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How can we simply apply the bohr model on a positronium as the positronium doesn't even have a nucleus or nucleus-like entity like in a hydrogen like atom. Moreover this implies that the first postulate of bohr's model is not applicable here as there is no corresponding electrostatic force to drive it in a circle due to absence of nucleus?..so are we considering the electrostatic force from the positron instead?

Simply saying that bohr's model is applicable here and that we can use reduced mass in the bohr radius formula to calculate the radius of the positronium doesn't make sense to me..

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    $\begingroup$ In as far as the Bohr model is applicable to anything it is true, nevertheless your objection. $\endgroup$ – my2cts Aug 14 at 9:55
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    $\begingroup$ Somewhat related: physics.stackexchange.com/questions/231913/… $\endgroup$ – PM 2Ring Aug 14 at 10:21
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    $\begingroup$ The Bohr model assumes that the whole nuclear charge is at (0,0,0), the center,because the nucleus is so much heavier than the electron that the center of mass of the system is the same within errors with the system of the nucleus at rest. For a positive negative equal mass charge the kinematics have to be taken into account . $\endgroup$ – anna v Aug 14 at 15:56
  • $\begingroup$ Did you study the Bohr model for hydrogen with a moving proton, where both proton and electron move around their center of mass? If not, you should start there. Positronium just has a different mass for the positively charged particle. $\endgroup$ – G. Smith Aug 14 at 16:30
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    $\begingroup$ The 2-body Coulomb problem is mathematically equivalent to a one-body problem with reduced mass plus a center-of-mass problem. There's nothing to understand beyond this and no need for a nucleus to be heavier than anything else. The location of the center-of-mass depends on the relative masses, and with the proton+electron system happens to coincide almost exactly with the position of the proton, but that's not required either by the physics of the problem or the math of the problem. $\endgroup$ – ZeroTheHero Aug 14 at 17:06

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