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Little Bang

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Electromagnetic spectrum

1. The annihilation of an electron and positron yields two photons, each with the energy of a single electron. That energy being 0.511 MeV.

2. Putting this value into the equation E = hv, where v = frequency, h is planck’s constant. V = E/h, This yields a frequency for the emitted photons of 124 EHz.

3. Is this value a coincidence, or could it have some hidden meaning?

 

ELECTROMAGNETIC SPECTRUM

Frequency Wavelength Energy

300 EHz 1 pm 1.24 MeV

30 EHz 10 pm 124 KeV

3 EHz 100 pm 12.4 KeV

300 PHz 1 nm 1.24 KeV

30 PHz 10 nm 124 eV

3 PHz 100 nm 12.4 eV

300 THz 1 um 1.24 eV

30 THz 10um 124 meV

3 THz 100um 12.4 meV

300GHz 1 mm 1.24 meV

30 GHz 1 cm 124 ueV

3 GHz 1 dm 12.4 ueV

300 MHz 1 m 1.24 ueV

30 MHz 1 dam 124 neV

3 MHz 1 hm 12.4 neV

300 KHz 1 Km 1.24 neV

30 KHz 10 Km 124 peV

3 KHz 100 Km 12.4 pev

300 Hz 1 Mm 1.24 peV

30 Hz 10 Mm 124 feV

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That's right.

 

Little Bang, the law of conservation of energy makes the two photons in total have the added energies of the electron and positron. Including kinetic energy. Try working it out with the energy-momentum 4-vector. If you could practically have the reaction occur with both leptons at zero momentum you'd be right about each photon's energy but try doing that for real, you'll run into a slight problem or two.

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The electron and positron had no motion relative to each other when annihilation occurred. With this in mind the calculation is correct.
Not only relatively to each other but also to your spectrometers!

 

Essentially, during the process they form a system called a positronium atom. If you send slow positrons onto still matter you'll see a line peaked fairly well around 511 keV from ortho-positronium but there's also para-positronium.

 

Another question, why is it that when a neutron decays the decay particles mass does not equal the mass of the neutron?

I'd say, more exactly, the product particles will never pop out at zero momentum, so count the kinetic energy and don't forget to count the neutrino in the products.

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