Little Bang Posted February 11, 2006 Report Posted February 11, 2006 Electromagnetic spectrum1. 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 SPECTRUMFrequency Wavelength Energy300 EHz 1 pm 1.24 MeV30 EHz 10 pm 124 KeV3 EHz 100 pm 12.4 KeV300 PHz 1 nm 1.24 KeV30 PHz 10 nm 124 eV3 PHz 100 nm 12.4 eV300 THz 1 um 1.24 eV30 THz 10um 124 meV3 THz 100um 12.4 meV300GHz 1 mm 1.24 meV30 GHz 1 cm 124 ueV3 GHz 1 dm 12.4 ueV300 MHz 1 m 1.24 ueV30 MHz 1 dam 124 neV3 MHz 1 hm 12.4 neV300 KHz 1 Km 1.24 neV30 KHz 10 Km 124 peV3 KHz 100 Km 12.4 pev300 Hz 1 Mm 1.24 peV30 Hz 10 Mm 124 feV Quote
Little Bang Posted February 12, 2006 Author Report Posted February 12, 2006 I am a little astonished that there have been no comments on this item. Quote
Qfwfq Posted February 13, 2006 Report Posted February 13, 2006 I hadn't seen it. Anyway, it can happen, don't take it bad. Are you sure the photons have both the exact same energy? What's the condition for this? Quote
Little Bang Posted February 13, 2006 Author Report Posted February 13, 2006 Q, the law of conservation of energy makes the photons have the energy of the electron, positron. It just seems strange that the digits of the calculated frequency would be the same as the energy of all those frequencys listed. Quote
Erasmus00 Posted February 14, 2006 Report Posted February 14, 2006 It just seems strange that the digits of the calculated frequency would be the same as the energy of all those frequencys listed. All that list does is change each side by a factor of ten. You could create such a chart for any number. -Will Quote
Qfwfq Posted February 14, 2006 Report Posted February 14, 2006 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. Quote
Little Bang Posted February 15, 2006 Author Report Posted February 15, 2006 The electron and positron had no motion relative to each other when annihilation occurred. With this in mind the calculation is correct. Another question, why is it that when a neutron decays the decay particles mass does not equal the mass of the neutron? Quote
InfiniteNow Posted February 15, 2006 Report Posted February 15, 2006 Another question, why is it that when a neutron decays the decay particles mass does not equal the mass of the neutron?deleted by InfiniteNow Quote
Qfwfq Posted February 15, 2006 Report Posted February 15, 2006 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. Quote
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