Positronium atom transition - energy of photon

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SUMMARY

The discussion centers on the energy transition of a positronium atom, which consists of an electron and a positron. When transitioning from the state n = 3 to n = 1, the energy of the emitted photon is calculated using the formula E = -13.6 eV x Z² / n². In the case of positronium, the effective atomic number Z is 0, leading to a calculated energy difference of approximately 6 eV for the transition. The reduced mass of the system, due to the equal masses of the electron and positron, plays a crucial role in determining the Rydberg constant for positronium.

PREREQUISITES
  • Understanding of atomic structure and quantum mechanics
  • Familiarity with the concept of reduced mass in two-body systems
  • Knowledge of the Rydberg formula and its application to hydrogen-like atoms
  • Basic proficiency in calculating energy levels of quantum systems
NEXT STEPS
  • Study the derivation of the Rydberg formula for hydrogen-like atoms
  • Learn about the concept of reduced mass and its implications in quantum mechanics
  • Explore the properties and behavior of positronium in quantum physics
  • Investigate photon emission and absorption processes in atomic transitions
USEFUL FOR

Students of quantum mechanics, physicists interested in atomic interactions, and anyone studying the properties of exotic atoms like positronium.

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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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