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Energy of photon emitted from Positronium transition

  1. Nov 7, 2005 #1
    Let's say I have a Positronium atom (an atom formed by an electron and a positron), and that this atom makes a transition from an n=3 state to an n=1 state. How do I find the energy of the photon that gets emitted during this transition?
    I've tried to use the Lyman series to find the wavelength. This looks like the following
    [tex]
    \frac{1}{\lambda}=R\left( \frac{1}{1^2}+\frac{1}{n^{3}} \right)
    [/tex]
    Setting n=3 I get
    [tex]
    \frac{1}{\lambda}=\frac{8}{9}R
    [/tex]
    If energy is E=vh, R=1.097e7m^-1, h=4.14e-15eV*s, and c=2.998e8m/s then
    [tex]
    E=vh=\frac{c}{\lambda}h=\frac{8}{9}Rch=12.10eV
    [/tex]
    I am looking for the answer E=6 eV, so I am pretty sure that I am wrong. Can anyone tell me how to do this correctly?
     
    Last edited: Nov 7, 2005
  2. jcsd
  3. Nov 7, 2005 #2

    Doc Al

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    Staff: Mentor

    reduced mass correction

    R, the Rydberg constant, depends on the mass of the orbiting particle. For ordinary hydrogen, where the electron orbits the massive proton, m is just the mass of the electron. But for positronium, the electron and positron have the same mass. To model the electron as orbiting the positron, you must use the reduced mass in calculating R. (The reduced mass = half of the mass of the electron.)
     
  4. Nov 8, 2005 #3
    So then does [tex]R= \frac{em_e}{8\epsilon_{0}^{2}h^3c}[/tex] in all cases?
     
  5. Nov 8, 2005 #4

    Doc Al

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    Staff: Mentor

    I suppose that by definition R equals that expression. But it assumes that the nucleus is fixed, which is a reasonable assumption for hydrogen where the nucleus is massive compared to the electron. But it won't do at all for positronium: you must use the more accurate expression that replaces the mass of the electron with the reduced mass of the electron-nucleus system.
     
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