Positronium atom transition - energy of photon

AI Thread Summary
The discussion revolves around the energy transition of a positronium atom, which consists of an electron and a positron. When transitioning from state n = 3 to n = 1, the energy of the emitted photon is questioned, particularly how it relates to the atomic number Z. It is clarified that for positronium, Z is effectively 0, and the energy levels are calculated using the reduced mass of the system. The calculated energy levels indicate that the photon energy emitted during this transition is approximately 6 eV. The complexities of the two-body problem and the Rydberg constant's dependence on mass are also highlighted.
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Homework Statement


Positronium is an atom formed by an electron and a positron. It is similar to the
hydrogen atom, with the positron replacing the proton. If a positronium atom
makes a transition from the state n = 3 to a state with n = 1, the energy of the
photon emitted in this transition is closest to

The Attempt at a Solution



what will my proton number Z be in this case?

E = -13.6eV x Z2 / n2

but a hydrogen atom's Z = 1, what about the positronium atom? 0?

state n=3 has -1.511 eV while state 1 has -13.6 eV.

but the answer is 6 eV? how is that obtained? i have no idea . thanks for help!
 
Physics news on Phys.org


The positron and electron have about the same mass. The electron does not revolve around the positron, but both revolve around their common centre of mass. This two-body problem can be treated as the motion of a single body with the reduced mass: mp*me/(mp+me) which is half the mass of the electron. The Rydberg constant is proportional to the mass of the electron: Now it is halved.

ehild
 


ah isee.. thank you very much ehild!
 
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