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Beta - and Beta + Decay (Q-Value)

  1. May 30, 2015 #1
    1. The problem statement, all variables and given/known data
    1) Calculate the Q-value for the electron emission beta decay of Co-60 *

    2) Calculate the Q-value for the positron emission beta decay of Mg-23 *

    * (both correct to 6 significant figures)

    Atomic Masses (u)
    electron = 0.0005485803
    Co-60 = 59.9338222
    Ni-60 = 59.9307906
    Mg-23 = 22.9941249
    Na-23 = 22.98976967

    Atomic Number
    Co-27
    Ni-28
    Mg-12
    Na-11

    2. Relevant equations

    Q = Σmic2-Σmfc2
    Beta decay (electron emission) = X ---> Y(Z+1) + -1e + v_
    where v_ = anti-neutrino
    Beta decay (positron emission) = X ---> Y(Z-1) + +1e + v
    where v = neutrino

    3. The attempt at a solution
    1 (Co-60)

    Q = (59.9338222 - (59.9307906+0.0005485803))*c2
    = 59.9338222 - 59.9313392
    = 0.0024830 * 931.5020
    = 2.3129
    = 2.31290 MeV (correct to 6 significant figures)

    2 (Mg-23)

    Q = (22.9941249 - (22.98976967+0.0005485803))*c2
    = 22.9941249-22.99031825
    = 0.0038066*931.5020
    = 3.54580 MeV (correct to 6 significant figures)

    Q is positive in both cases. Q is the energy that is liberated in the reaction. Am i correct in neglecting and mass that the anti-neutrino and neutrino have?
     
  2. jcsd
  3. May 31, 2015 #2

    BvU

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    Neglecting the neutrino mass is fine.
    But e.g. here I find a different decay energy. Can you spot the difference and understand why ?
     
  4. May 31, 2015 #3
    The Co-60 value looks similar, but the Mg-23 does not. Your reference says that Mg-23 decays via electron capture and not positron emission beta decay.

    Other than that I'm not really sure what the difference is, because Q = Σmic2-Σmfc2 and I've got all the masses in the reaction.
     
    Last edited: May 31, 2015
  5. May 31, 2015 #4

    BvU

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    Co60:
    Did I give the right link ? It's on table4. To me it does not look similar at all: 2.824 versus 2.313 ?

    Are you sure you have to deduct the electron mass ? That means you consider it as being created from the reaction energy [edit] which is correct. But how about your statement "have all the masses"?
    Checking here might set you thinking about what the given data stand for: the nuclear masses or the atomic masses ?
     
  6. May 31, 2015 #5
    I've been using the atomic masses to calculate the Q-value. The second link you sent suggests I should use the nuclear mass. I'm a bit confused now
     
  7. May 31, 2015 #6
    Q = Σmic2-Σmfc

    Q = Σmic2 - Σ mfc2

    mi = atomic mass of element - number of electrons = mass of nucleus

    mf = (atomic mass of element - number of electrons) + mass of one electron = mass of nucleus + one electron

    Is these the correct masses I should be using? I thought m was the atomic mass, but from what I gather in what you've sent me it's the mass of the nucleus (nuclear mass).
     
  8. May 31, 2015 #7
    Q = ( 59.9338222 - 0.0148116681 ) - ( 59.9307906 - 0.0153602484 + 0.0005485803 ) * c2
    = 0.0030315 * 931.5020
    = 2.8238
    = 2.82380 MeV (correct to 6 significant figures) :smile:

    I'll have to do the same for the other one too!

    Just a quick question. I've been asked to calculate this to 6 sig figs. 0.0030315 * 931.5020 = 2.823848313, however because we are multiplying we have to round to the .0000 because of the 931.5020. This makes the answer 2.8238, and because the answer is wanted to 6 sig figs it = 2.82380 MeV. Is my reasoning correct here?
     
  9. May 31, 2015 #8

    BvU

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    I wouldn't subtract 27 me and 28 me on each side and then add one me on the Ni again,....
    Besides the neutrino we also ignore the difference in total binding energy (check the wikipedia article -- looks like that's ok to do).

    Subtracting the atomic masses as give leaves only 5 digits. Were the atomic masses given in the exercise ? Then the exercise itself leaves you a digit short !
    If you quote 0.0028238 GeV/c^2 you could argue that's seven digits.

    [edit] the 931.502 is 6 digits, so that's ok. -- Note that it's about relative errors when multiplying !
    But I found 931.4941 ?!

    (have to run!)
     
    Last edited: May 31, 2015
  10. May 31, 2015 #9
    Yeah, the atomic masses were given as:

    electron = 0.0005485803
    Co-60 = 59.9338222
    Ni-60 = 59.9307906

    Conversions were given as well
    1 eV = 1.602189E-19 Joules
    c = 2.99792458E+08 m/s
    1 amu = 1.660566E-27 kg
    1 amu = 931.5020 MeV

    There's no way I'm converting amu to kg, calculating joules, converting to MeV. No thank you!

    59.9338222 - 0.0148116681 = 59.9190105 ( 9 sig figs )
    59.9307906 - 0.0153602484 + 0.0005485803 = 59.9159790 ( 9 sig figs )

    59.9190105 - 59.9159790 = 0.0030315 (5 sig figs)

    0.0030315 (5 sig figs) * 931.5020 (7 sig figs) = 2.8238 (5 sig figs) = 2.82380 (6 sig figs)
     
  11. May 31, 2015 #10
    (Mg-23)

    Q = [ (22.9941249 - 0.0065829636) - (22.98976967 - 0.0060343833 + 0.0005485803) ]
    = 0.0032580 * 931.5020
    = 3.0348
    = 3.03480 MeV (correct to 6 sig figs)

    Your link suggests Mg-23 decays via electron capture and the decay energy = 4.057 MeV (http://atom.kaeri.re.kr/ton/nuc4.html)
     
  12. May 31, 2015 #11
    This link, at the bottom of the page, says that B+ decay of Mg-23 = 3.0348 MeV

    http://crete.homeip.net/show_nuclide/120023/ [Broken]
     
    Last edited by a moderator: May 7, 2017
  13. May 31, 2015 #12

    BvU

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    A difference of 2 me .

    Again, the wikipedia link explains why the 3.0348 MeV is the right answer -- in 5 digits. Really
    is bluffing: you don't know it's a zero at the end. But as I stated: not your fault.
     
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