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Fusion Help

  1. Mar 17, 2008 #1
    I need a bit of help here, as part of a class project i was trying to explain how fusion, binding energy, reactors etc work. I always like to go a bit more indepth than the syllabus requires, and i wanted an explanation of how D-T fusion worked mathematically, to clearly show how the binding energy becomes output energy - however, whenever i do the calculations, i don't get the same net energy output as I find online, 17.6 MeV. I always end up with about 17.1.
    I tried to compensate for the initial energies, relativistic masses and the coulomb barrier, but I'm still not a huge deal closer. Can anyone give me any clues as to what i may be missing? If nothing is obvious i'll add a brief of summary of the calculations i did.
    Thanks, Tom.
  2. jcsd
  3. Mar 17, 2008 #2


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    Well there is no way that we can see what misstake you made if you not show us your work.
  4. Mar 17, 2008 #3
    Ok, I'll give a brief summary, i was just wondering if I'd missed something face slappingly obvious. this is before i started trying to correct it with extra bits, they didn't help a huge amount.

    Deutrium 2.013004633
    Tritium 3.01550062
    Total 5.028505253

    Helium 4.0015
    Neutron 1.008664917
    Total 5.010164917

    Missing amu 0.018340336
    In kg 3.04548E-29
    E = MC2
    energy 2.73714E-12

    MeV 17.08
  5. Mar 17, 2008 #4


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    He-4 atomic mass is 4.00260323 amu, there is your error.
  6. Mar 17, 2008 #5
    i used the mass of an alpha particle, which explains the difference, because the electrons shouldnt be involved in a nuclear reaction should they?
  7. Mar 17, 2008 #6


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    then you must remove the electrons from the deuterim and tritium also. Remember that what is listes are ofte the ATOMIC masses. So one must be careful...and cosistent.

    Yes you are right, in reality (in a fusion reactor) only nuclei takes part in the reaction.

    Also: Post course work / homework questions in appropirate home work forum, this forum is not for those kind of questions which can be read here: https://www.physicsforums.com/showthread.php?t=171607
  8. Mar 17, 2008 #7
    I did remove the electrons from them too, sorry, i should have made that clearer. I didn't think to post it in the homework area because it isn't part of my syllabus, this is more of an outside concern really.
  9. Mar 17, 2008 #8


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    ok in that case:

    nuclear masses before=
    5.02905362 amu

    Nuclear masses after=
    5.010170986 amu

    gives 17.58906 MeV
  10. Mar 17, 2008 #9
    Right, thanks, that's close enough, and when i use those values it works with my calculations. my values for atomic mass must have been slightly off... for future reference, where do you go to find yours?
    Thanks for your help, Tom.
  11. Mar 17, 2008 #10


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  12. Mar 18, 2008 #11
    Thanks so much for your help, that website is really usefull! just out of interest though, can you think of anything that might account for the slight discrepancy we're still getting?

    My calculations say that if the particles were travelling at about 2.5 million metres per second at the start of the reaction that would be enough extra energy to account for it... that sounds like a reasonable enough speed for high energy particles in a 100,000,000 K fusion reactor... am i on the right lines here maybe?
    Last edited: Mar 18, 2008
  13. Mar 18, 2008 #12


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    1. The source you have ARE already listing nuclear masses and you substract the electron mass again..

    2. Lack of accurateness, use ALL significant figures that you can. 17MeV is a very very small unit (3% of electron mass, and one electron is 1/1800 of a nucleon), so be careful!

    3. The source you used have listed totaly wrong numbers for atomic masses.

    With the numbers I used, I got 17.6MeV, which is what the answer tells you.
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