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Positronium transition, energy of photon emitted

  1. Dec 16, 2004 #1
    What is the energy of the photon emitted when a positronium atom goes from the n=3 state to the n=1 state?

    Edit:
    Nevermind, I figured it out.

    [tex]
    E_n = -\frac{\mu}{2{\hbar}^2} (\frac{Ze^2}{4\pi\epsilon_0})^2 \frac{1}{n^2}
    [/tex]

    For hydrgen, the effective mass is approximately the mass of the electron.
    [tex]\mu=m_e[/tex]

    However, for positronium,
    [tex]\mu=\frac{m_pm_e}{m_p+m_e}=\frac{m_em_e}{m_e+m_e}=\frac{m_e}{2}[/tex]

    So the length of the energy is only half that of the hydrogen atom.

    Therefore,
    [tex]E_3-E_1=(\frac{1}{3^2}-1)(-6.8 eV)=-\frac{8}{9}*-6.8 eV = 6.04 eV[/tex]


    grn. 31.
     
    Last edited: Dec 16, 2004
  2. jcsd
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