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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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It is speculative but a statement can be made that it is impossible to 100 percent annihilate a electron and positron if there is a charge parity in that area . 99 percent yes but not 100 percent do the the conservation of energy . We on earth are about 300 K volts negative relative to space . Stated another way we have a 300 k volt sky charge meaning earth has a charge parity of 300 K volts compared to neutral space . In order to bring one positron and one electron from space down to earth a voltage barrier of 300 K volts must be past . With a strict application of e=mc^2 the mass of the electron must go up and the positron down . It is a very small difference but it is there just the same . A 100 percent annihilation of a electron / positron pair would be a violation to the conservation of energy as the masses would not be equal . This allows wiggle room for a ash left over from annihilation . Two photons and one very low energy atom of positronium . It's mass would be in the neighbourhood of a neutrino so good luck in detecting it . Just the same it should theoretically be there . I call the left over ash a VEPS particle , quasi virtual positronium who's properties change with the charge parity of a given area . Add 1 M volt per VEPS vacuum and pair production returns us to the original electron and positron . The notion that this type of low energy positronium would have a radius of two times hydrogen sounds about right if our space , vacuum , is composed of quasi virtual positronium left over from a hotter universe.

John

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This looks completely wrong; unless it's based on some physics which I'm not familiar with. –  Ali Jul 15 '13 at 2:45
    
This is complete nonsense. I particle physics mass always refers to rest mass not "relativistic mass." Rest mas is an invariant. Electron/positron masses are not at all affected by a background EM field and neither is the annihilation reaction which is a local process. This person does not understand basic relativity. I wonder how he thinks that the LEP experiment worked! By the way a very low mass electron/positron composite would be easy to detect... we can detect neutrinos after all! –  Michael Brown Jul 15 '13 at 10:41
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