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Q-value in beta minus decay does not match.

  1. Jan 9, 2012 #1

    I'm wondering if someone could help me out finding the Q-value fom a beta minus decay.

    It's a
    90Y → 90Zr + e

    All sources I have says that the energy released is 2,28 MeV, but when i calculate i get 1,78 MeV

    So far, I've looked up the mass of the 90Y and the Zr90 and the electron and done following:


    According to my databook

    m(90y) = 89,9071519u = 1,49294*10^-25 kg
    m(90Zr) = 89,9047044u = 1,4929*10^-25 kg
    m(e) = 9,10938*10^-31 kg

    Giving a mass difference at 3,15323*10^-30 kg

    resulting in a Q value of Q=2,83399*10^-13 J = 1,76883 MeV

    If you can tell me what im doing wrong, I would appreciate if you would let me know.

    Thanks in advance.
  2. jcsd
  3. Jan 9, 2012 #2


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    I would just point out that the difference between your answer and the expected answer is about 0.5 MeV, i.e. the mass of the electron.
  4. Jan 9, 2012 #3


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    Staff: Mentor

    Hint: isotopic mass tables list the masses of complete atoms (including electrons), not bare nuclei.
  5. Jan 9, 2012 #4
    I finally got the right result!

    Thanks a lot :)

    I tried only calculation with masses of the mother- and the daugther nucleus, without the β particle itself, and that gave me the 2,28 MeV i was looking for.

    But, I'm still confused. How can i get it right, when im ignoring a product? if
    Q = (m(product) - m(reactants) )*c^2

    Am i just allowed to ignore the elektron?
    Or is the electron actually a part of the daughter nucleus as n→p+e?
  6. Jan 9, 2012 #5


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    Is it really the case that people calculate their nuclear reactions nowadays in terms of Joules and Kilograms?
  7. Jan 9, 2012 #6


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    Staff: Mentor

    Hint #2: How many electrons does a Y atom have, and how many electrons does a Zr atom have?
  8. Jan 9, 2012 #7
    My teacher prefer that we use SI-Units, as we will be sure that the results are in SI-Units.
    Also, I don't know what county the majority of the users of this forum is, but I live in Denmark, and here we use the metric system (Don't know if that even matters here).

    The Y has 39, and the Zr has 40. Is the extra electron from the Zr the β? If so, how is the Zr atom stable?
  9. Jan 9, 2012 #8


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    Gold Member

    1) What Bill_K is referring to is that nuclear reactions (in my experience, and apparently his) are computed using amu and mev directly. Going to joules and kilograms seems lunatic.

    2) Do you think, when Y decays by beta, and emits a high energy electron which escapes, the result is a neutral Zr or an ionized Zr? Account for this and all will balance.
  10. Jan 9, 2012 #9


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    "Something is rotten in Denmark."
  11. Jan 10, 2012 #10
    Hmmm... The calculation in diferent unit is not a matter here, as my calculator is able output the result in the right units and aswell able to convert from kg to amu, Joule to eV amd so on.

    But I have now calculated the whole thing over again, with all taking the masses og the nucleus and removing electrons and getting the right energy, and seems to work. But Zr also seems to be taking an electron from it's surroundings, but then again, I don't see any reason why that couldn't happen.
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