# Is the minimum radius of a positronium system of the order of compton wavelength or less than that?

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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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.
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 $m_e\cdot m_e/(m_e+m_e) = m_e/2$.