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Fission reaction & Q-value

  1. May 30, 2015 #1
    1. The problem statement, all variables and given/known data
    Fission Reaction: Pu-239 + gamma --> Sr-92 + ? + 3n

    Q-Value = ? (Answer in MeV, correct to 5 significant figures)

    Atomic Masses (u)
    Alpha Particle = 4.00150618
    He-4 = 4.0026032497
    Pu-239 = 239.0521565
    Sr-92 = 91.911030
    n (neutron) = 1.00866501

    2. Relevant equations

    Q-Value = Q = E = Δmc2 = (mi - mf)c2

    3. The attempt at a solution

    Q = (239.0521565 + gamma) - (91.911030 + 4.0026032497 + (3*1.00866501)) * c2
    = (239.0521565 + gamma) - (98.939628)
    Q = gamma = -140.1125285 * 931.5020
    = -130515.1
    = - 130520 MeV (correct to 5 significant figures)

    Two Questions
    1) I've seen examples where He-4 = 4.0026032497 is used in this calculation instead of the Alpha Particle = 4.00150618. Why is this? I know He-4 is missing it's two electrons, therefore it's mass is the equivalent of the alpha particle, but how come He-4 atomic mass ≠ alpha particle. And why should I use the atomic mass of He-4 in this calculation instead of the atomic mass of the alpha particle?

    2) Is Q negative because it represents the amount of energy which must be provided for the reaction to occur? I.e. A gamma ray with 130520 MeV must be provided to split a Pu-239 atom into Sr-92 + He-4 + 3n
     
  2. jcsd
  3. May 30, 2015 #2
    I'm a bit confused as to why you've introduced an alpha into your solution? Looking at the problem statement there are only the Pu, gamma, Sr, and 3n.
     
  4. May 30, 2015 #3
    Pu-239 decays via alpha decay.

    so Pu-239 + gamma --> Sr-92 + ? + 3n

    If Pu-239 decays via alpha decay then ' ? ' = He-4

    Pu-239 + gamma --> Sr-92 + He-4 + 3n

    Sorry, that was the first part of the question but I figured that out from a quick google search.
     
  5. May 30, 2015 #4
    To answer your question on the mass difference between He-4 and an alpha, He-4 has two electrons (which are on the order of 0.0005amu each) and an alpha has none - an alpha is just a He-4 nucleus.
     
  6. May 30, 2015 #5
    Your Q value should be positive, it is the amount of energy released from the reaction. When I read the question as set up what I see is gamma induced fission of Pu resulting in Sr, 3n, and a lot of energy. The fact that Pu-239 decays via alpha emission doesn't factor in. So what I would set up to solve is ((mass Pu)-(mass Sr + mass 3n))c^2
     
  7. May 30, 2015 #6
    Sorry, the full question states:

    Complete the following fission reaction, and then calculate the Q-value for the reaction: Pu-239 + gamma --> Sr-92 + ? + 3n

    I'm pretty sure that means Pu-239 + gamma --> Sr-92 + He-4 + 3n

    I thought because the energy of the gamma ray is the only unknown then Q = energy of the gamma ray = 130520 MeV

    Because the energy of the gamma ray is the amount of energy that has to be provided for the reaction of Pu-239 + gamma --> Sr-92 + He-4 + 3n to occur, this means gamma = -130520 MeV
     
  8. May 30, 2015 #7

    Astronuc

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    The OP indicates gamma-induced fission (as opposed to neutron-induced fission). The ? should be a fission product, not an alpha particle. The sums of Zs and As on the leftside of the equation should equal the sum of the Zs and As. The gamma ray has no charge or mass. Z(Pu) = 94, Z(Sr) = 38. Neutrons have no charge, but atomic mass, A = 1.
     
  9. May 30, 2015 #8
    The Q value is a measure of the energy released in a reaction. So in this case it is the energy released in a Pu+gamma reaction (gamma induced fission of Pu). A positive Q value means the reaction is exothermic (which fission very much is), a negative would indicate the reaction was endothermic.
     
  10. May 30, 2015 #9

    SteamKing

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    This is why the PF HW forum rules ask posters to include the full question when filling out the HW template. Otherwise, a lot of time is wasted in going back and forth trying to make sense of what is being asked.

    Also, the fission reaction:

    Pu-239 + gamma --> Sr-92 + He-4 + 3n

    needs examination. Has anyone tried to balance the left hand side with the right hand side? What's the atomic number of plutonium? Of strontium?

    A gamma ray with an energy of 130,000 MeV is an extraordinary event. You only get about 200 MeV from the fission of a single atom of U-235.
     
  11. May 30, 2015 #10
    Pu-239 ---> Sr-92 + Ba-147 + 3n

    Z
    Pu = 94
    Sr = 38
    Ba = 56

    Ba-147 = 146.93399 u

    Q = E = Δmc2

    = (239.0521565 - (91.911030 + 146.93399 + 3.02599503) * c2
    = - 2.81886 * 931.5020
    = -2625.77
    = -2625.8 MeV (correct to 5 significant figures)

    I think I've calculated it correctly now. I'm not sure about the negative/positive value. My notes say that 'if Q is negative, it represents the amount of energy which must be provided in order for the reaction to occur.'
     
  12. May 30, 2015 #11

    SteamKing

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    Did you add up the atomic masses on both sides of the fission equation?
     
  13. May 30, 2015 #12
    Closer. However, you appear to have created mass with your fission products.
     
  14. May 30, 2015 #13
    Left hand side = 91.911030 + 146.93399 + 3.02599503 = 241.8710150
    Right hand side = 239.0521565 + gamma

    Q = E = Δmc2
    = (minitial - mfinal) * c2

    (239.0521565 + gamma)*c2 ---> 241.8710150*c2

    222,677.5619 - 225,303.3342
    = - 2625.77 MeV
    = - 2625.8 MeV (correct to 5 significant figures)
     
    Last edited: May 30, 2015
  15. May 30, 2015 #14

    SteamKing

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    You only started with on atom of Pu-239, on the left hand side. How did you get "Left hand side = 91.911030 + 146.93399 + 3.02599503 = 241.8710150"?
     
  16. May 30, 2015 #15
    Sorry, meant to be

    right hand side = 91.911030 + 146.93399 + 3.02599503 = 241.8710150
    left hand side = 239.0521565 + gamma
     
  17. May 30, 2015 #16

    SteamKing

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    Your Q value is still about an order of magnitude too large.
     
  18. May 30, 2015 #17
    How could this be? I've used the correct units

     
  19. May 30, 2015 #18

    Astronuc

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    The atomic masses must be equal. 239 + 0 (gamma) = 92 + ? + 3.
     
  20. May 30, 2015 #19

    SteamKing

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    Generally, fission reactions are exothermic, greatly so. Why else would you want to build a nuclear plant, if you had to keep shoving all this energy into it to split atoms?

    Also, fission reactions involving uranium and plutonium evolve pretty much the same energy per atom:

    http://www.kayelaby.npl.co.uk/atomic_and_nuclear_physics/4_7/4_7_1.html
     
  21. May 30, 2015 #20
    I assumed because this is a question about mass-energy equivalence, q-value / decay energy that the difference between products and reactants is what is needed to be input for the reaction to occur.

    mf*c2 does not always equal mi*c2 because of the different binding energies of the product nuclei.

    If Q = ∑mfc2 - mic2

    (239.0521565)*c2 - 241.8710150*c2

    -2.8188585 * 931.5020
    = -2625.77 MeV
     
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