The allowed radii and energies of positronium

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SUMMARY

The allowed radii and energies of positronium, a hydrogen-like atom consisting of a positron and an electron, are defined by specific equations. The radii are calculated using the formula r = (4n²πε₀ħ²)/(μe²), where μ is the reduced mass of the system. The energies are determined by the equation Eₙ = - (μe⁴)/(2n²(4πε₀)²ħ²). Calculated values for allowable radii and energies include rₙ = n²(1.05835 × 10⁻¹⁰ m) and Eₙ = (1/n²)(-6.8032 eV).

PREREQUISITES
  • Understanding of quantum mechanics principles
  • Familiarity with the concept of reduced mass
  • Knowledge of fundamental constants such as ε₀ and ħ
  • Ability to perform calculations involving energy levels and atomic radii
NEXT STEPS
  • Study the derivation of the reduced mass formula in quantum systems
  • Explore the implications of quantum mechanics on atomic structure
  • Learn about the differences between positronium and hydrogen in terms of energy levels
  • Investigate applications of positronium in particle physics and materials science
USEFUL FOR

Students and researchers in physics, particularly those focusing on quantum mechanics, atomic structure, and particle interactions. This discussion is beneficial for anyone studying the properties of positronium and its applications in advanced physics.

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Homework Statement



Positronium is a hydrogen-like atom consisting of a proton and an electron revolving around each other. Find the allowed radii and energies of the system.

Homework Equations



See solution attempt.

The Attempt at a Solution



The allowed radii of the positron/electron system are given by r = \frac{{4{n^2}\pi {\varepsilon _0}{\hbar ^2}}}{{\mu {e^2}}} where \mu = \frac{{{m_{positron}}{m_e}}}{{{m_{positron}} + {m_e}}} and n={1,2,3,...}

The allowed energies of the positron/electron system are given by {E_n} = - \frac{{\mu {e^4}}}{{2{n^2}{{(4\pi {\varepsilon _0})}^2}{\hbar ^2}}}

It just seemed suspiciously simple. Is this all they're really asking me to state? I mean, other than calculating it all out and leaving it as a function of n times a constant.
 
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Just in case that was what I needed to do, I calculated the following allowable radii and energies, using the reduced mass:

\begin{array}{l}<br /> {r_n} = {n^2}(1.05835 \times {10^{ - 10}} m)\\<br /> {E_n} = \frac{1}{{{n^2}}}( - 6.8032eV)<br /> \end{array}
 

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