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KeV, MeV and GeV....


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#1 YT2095

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Posted 19 January 2008 - 05:40 PM

what the idea behind these and particle interactions?

for instance the annihilation energy of a Positron and an Electron is 511KeV.
well that`s Nice and everything but...
since I have these exact antimatter anihilations going on in my Lab 24/7, why am I not seeing 50+ inch arks every time it happens?

also, since 511KV is reasonably easy to attain with a little time and effort, would that mean that If I ripped an arc at at this voltage I would be making antimatter?

what exactly IS the relationship between this voltage and Particles?

in Real terms please!
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#2 ydoaPs

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Posted 19 January 2008 - 05:48 PM

eV is a unit of energy, not potential difference.
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#3 YT2095

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Posted 19 January 2008 - 05:55 PM

so 1eV does not equal 1 Volt then?
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#4 timo

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Posted 19 January 2008 - 06:17 PM

No. 1 eV is the amount of kinetic energy of an electron that has travelled through a potential difference of 1 V (assuming it was at rest before). For elementary particles, think of eV as just being a different measurement scale than joule (like meter vs. centimeter) not as something related to voltage or electric charge. Sidenote: It's keV, not KeV.
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#5 swansont

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Posted 19 January 2008 - 06:28 PM

And you'd need twice the 511 keV to make antimatter, since you have to create an e-e+ pair to conserve charge and lepton number. And realistically a little more than that, so that the particles have some KE and can separate.
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#6 YT2095

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Posted 19 January 2008 - 06:53 PM

so a 1.1MV Tesla coil firing off will make antimatter?
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#7 John Cuthber

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Posted 19 January 2008 - 08:33 PM

No it won't.
The potential difference is large enough but an electron set free will be acelerated over a short distance before it hits an air molecule. It won't get to cover the whole distance from 1.1MV to ground so it won't get that much energy.
You would need to put the voltage across a vacuum tube to get a high enough energy to generate antimatter. Even then you would need to get the right sort of collision to get antimatter.
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#8 Ozone

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Posted 20 January 2008 - 04:00 AM

Good evening, all.

Think of eV as E as in E=mc^2:

the mass of an electron is appx. 9.109E-31 kg, therefore:

E=(9.109E-31 kg)(3E8 m/s^2)=8.1981E-14 kg.m^2/s^2 = 8.1981e-14 N (or J).

Since 1eV = 1.602E-19 J, 9.1981E-14 J/ 1.602E-19 eV/J = 511741.57 eV

or:

0.511...MeV. (a proton is appx. 938 MeV)

When a photon meets or exceeds an energy of ~1.022 MeV, it is in the range where pair production may occur. When this happens, the photon converts into an electron:positron pair. These annihilate to yield two photons of 0.511 MeV, which are emitted at 180, relative to one another.

This phenomena is readily observed with gamma spectrum analysis of 60Co.

This is an amazing proof to see in the lab. Mass and energy are perfectly conserved and convertable *drools*.

This is almost as cool as seeing a muon take minutes to decay when at near relativistic velocities.

Cheers,

O3
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