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Old 02-11-2006   #1 (permalink)
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Electromagnetic Energy

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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Old 02-12-2006   #2 (permalink)
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Re: Electromagnetic Energy

I am a little astonished that there have been no comments on this item.
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Old 02-13-2006   #3 (permalink)
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Re: Electromagnetic Energy

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?


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Old 02-13-2006   #4 (permalink)
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Re: Electromagnetic Energy

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.
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Old 02-13-2006   #5 (permalink)
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Re: Electromagnetic Energy

Quote:
Originally Posted by Little Bang
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
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Old 02-14-2006   #6 (permalink)
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Re: Electromagnetic Energy

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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Old 02-14-2006   #7 (permalink)
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Re: Electromagnetic Energy

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?
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Old 02-14-2006   #8 (permalink)
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Re: Electromagnetic Energy

Quote:
Originally Posted by Little Bang
Another question, why is it that when a neutron decays the decay particles mass does not equal the mass of the neutron?
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Last edited by InfiniteNow; 02-15-2006 at 12:02 PM.. Reason: Trying to speak of an issue I cannot fully articulate.
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Old 02-15-2006   #9 (permalink)
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Re: Electromagnetic Energy

Quote:
Originally Posted by Little Bang
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.

Quote:
Originally Posted by Little Bang
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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