Since from electron-positron annihilation energy and uncertainty principle,the minimum radius of positronium comes out as half of the compton radius.
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After annihilation there is no electron and positron in a new (lower) bound state, thus there is no positronium. Positronium decays into several photons, that's it. The photon wavelength has nothing to do with the "new positronium state" since there is no positronium in the final state. |
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No, the positronium radius is of the order of the Hydrogen atom radius (Bohr radius), $$ a_0 = \frac{4 \pi \varepsilon_0 \hbar^2}{m_{\mathrm{e}} e^2} = \frac{\hbar}{m_{\mathrm{e}}\,c\,\alpha} $$ which is longer than the Compton wavelength of the electron because of the extra factor $1/\alpha=4\pi\varepsilon_0\hbar c/e^2\sim 137.036$, the inverse fine structure constant. This factor is usually not considered to be "of order one" in these considerations. The Compton wavelength is "something in between" the size of the atom (or positronium) and the size of the nucleus. If you want it more accurately, the positronium is (almost) exactly 2 times larger than the Hydrogen atom – when it comes to the distances between the two charged particles in it – and its binding energy is 1/2 of the hydrogen value. That's because the "reduced mass" of the positronium problem is $m_e\cdot m_e/(m_e+m_e) = m_e/2$. |
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