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Nuclear yield math

  1. Aug 26, 2007 #1
    Greetings---I would like to fool around with numbers, finding the fusion and/or fission yield (in MeV) for various combinations of elements. I can do the simple ones involving Hydrogen and it's isotopes, but figure I'll run into trouble with more complex nucleuses. Can anyone point me in right direction?
  2. jcsd
  3. Aug 26, 2007 #2
    The chart of the nuclides on the National Nuclear Data Center website gives all the mass defects for each isotope, if you have that it's just a matter of finding the difference in the reactants and the products and converting to energy. If you're determined, you could make a data base for istopes you're interested in and write a small program to find the resulting energy.

    Here's the chart: http://www.nndc.bnl.gov/chart/
  4. Aug 26, 2007 #3
    WOW----way cool!!! However, I don't understand most of it. Please explain these two:

    E (level) (MeV)

    tiny triangle(MeV) (is that triangle called "delta" meaning "change"??)

    Not sure I understand what "mass defect for each isotope" means? Change in mass of it compared to the pure element??

    Could you possibly work through an easy fusion example, showing me your work?
  5. Aug 26, 2007 #4
    I'm not entirely sure about E (level) (MeV) since I've never had to use it, but my best guess is that it is the energy of the excited nucleus. Just as electrons can be in excited levels, the nucleuons in the nucleus can as well.

    Yes, that triangle is a delta and does signify change, but again I'm not sure of what. At first I assumed it was the mass defect, but if you calculate it yourself, you'll see that's wrong. Here is the formula for calculating mass defect

    [Z(mass Hydrogen) + (A - Z)(mass of neutron) - (mass of isotope with atomic number Z and total nucleons A)] * 931.5

    Where masses are in amu and the constant is [MeV/amu]. I got the masses from Fundamentals of Nuclear Science and Engineering by Shultis and Faw, but you can find them here: http://physics.nist.gov/cgi-bin/Compositions/stand_alone.pl?ele= A neutron's mass is 1.008665 amu, since that isn't listed there.

    Notice that the mass defect is given in terms of energy, this can also be called the binding energy. Finding the difference in binding energy will give you the "Q value" (amount of energy released) of the reaction.

    Here's an example of fusion of deuterium (Hydrogen with one neutron) and tritium (Hydrogen with two neutrons), the fuel that is to be used in ITER.

    The reaction is:
    D + T => He + n

    Using the equation above, you can find the mass defects are
    D => 2.224587 MeV
    T => 8.481877 MeV
    He => 28.29586 MeV

    Now, the binding energy of the product minus the binding energy of the reactants.
    BE(4He) - BE(2H) - BE(3H)
    28.29586 - 2.224587 - 8.481877 = 17.5894

    So, from one reaction of a deutron and triton, you get about 17.6 MeV, unfortunatly most of the energy goes to the neutron, and can't be immediatly converted to useful energy.

    Using the Q calculator, you can find this is a good answer
    You might be temped to use the calculator altogether, but I suggest you do a couple calculations on your own before using it.
  6. Aug 27, 2007 #5
    Candyman--I sure appreciate your help. But, I have been able to do the easy ones for some time. If you go to the PHYSICS forum, then down to "High energy, particle,....." I have a few posts. Go into the one called "Binding Energy--same question-different approach" In post #8 and #10 I had already done the stuff you showed me. However, I was glad you showed me anyways--happy to see you got same answers I did!!! lol

    I am thinking more along the lines of more complicated fusion. I am thinking along the lines of 2 Li6 to make 1 C12. Or maybe Carbon12---->Sulfer32=Titanium44. I can only imagine how many sheets of paper that one would take!! To throw another twist into it, start adding together parts that put the product "over the 60 hump" of the Iron Group on the Binding Curve, watching the energy go down again. Then, for a finale, try to fuse two huge atoms (U and Pu for laughs), and see the negative energy answer (hopefully!). This is the maddnes I would like to do.

  7. Aug 27, 2007 #6
    Ah, sorry. I thought it would be best to give simple examples if you hadn't done this before.

    I don't think it will be possible for you to fuse Pu and U, because there isn't any data about the products of such a reaction - as far as I know, anyway. You can use that Q calculator I linked to in my pervious post to calculate Q vales of reactions, if you know the reactants and the products.
  8. Aug 27, 2007 #7
    I am now attempting Li6 + Li6==>C12 I added the mass of 6 protons, plus the mass of 6 neutrons, then subtracted the total from the given mass of C12

    6(1.007276) + 6(1.008665)==>12.0107

    Defect was .084 expressed in energy that would be:

    .084 x 931.484=78.2445 MeV

    OK, that is the binding energy of the Carbon atom--divide that by 12 gives BE/nucleon, if you wish to see that.

    So, here's what if have now, in an attempt to get yield:

    78.2455 - BE(Li6) - BE(Li6) = Yield

    I think I know what to do, calculate the BE for Li, but thought I'd ask a few questions first:

    1.) Will there be a different result for yield depending on which of these you use??

    (a.) 12 individual pieces (6p and 6n) fused all at once, in one step
    (b.) Fusing 2 atoms of H3 for each of the 2 of the Li atoms
    (c.) Fusing 3 atoms of H2 for each of the 2 of the Li atoms
    (d.) Using method (b.) for one Li atom, and method (c.) for other Li atom

    My thinking is that (a.) will produce 78.2455 MeV for yield, but (b.), (c.), and (d.) all need to be broken down even more, to get the BE for Deuterium and Triterium. Am I on right track? Is there 4 different possibilities for yield in the Li6 + Li6 ==> C12???
    Last edited: Aug 27, 2007
  9. Aug 27, 2007 #8
    Hmmm, I'm not sure I can answer you 100%, but I'll tell you what I'm sure of.

    A) That's right, this is the basic definition of binding energy.
    B) Fusing two tritons will not yield lithium.

    C) and D) I'm guessing at your thinking here, but I'll tell you this right now, at some point, everything could have been all protons and neutrons. Now matter how you put them together, the mass defect will always be the same, so no matter what route of fusing you take, it is impossible that all the energy given off getting from point 1 to point 2 differs by the reactions.

    It works the same going backwards, say something fissions and those fission products decay into A, B, C and D; then if the same original type of atom fissions into A, B, C and D right off without having daughter products that decay, the same amount of energy has been given off.
  10. Aug 27, 2007 #9
    Quick note on (b.) I had a brain fart, and somehow thought Triterium had 3 protons. In fact, I see now I screwed up all of it. My mind was thinking Triterium had 3 protons, Deuterium had 2 protons, and Li had 6 protons. I was at work, what can i say!!! My bad!! :(
    Last edited: Aug 27, 2007
  11. Aug 28, 2007 #10
    The 78.24MeV is also wrong--I used the mass for Carbon12 off of the Periodical Table--which, as you know, is NOT the mass of C12, but the average mass of all the C isotopes. I will now retreat and regroup!!! lol :) :) :)
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