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Finding Q values of decays and finding distance where Coulomb Barrier = Q value

  1. Jan 21, 2012 #1
    1. The problem statement, all variables and given/known data
    224Ra --> 220Rn + [itex]\alpha[/itex]
    224Ra --> 212Pb + 12C
    224Ra --> 210Pb + 14C

    Calculate the Q-Values (in MeV) for these decays given their atomic mass excesses (in MeV) are

    88225Ra = 18.818 86220Rn = 10.604
    82212Pb = -7.557 [itex]\alpha[/itex] = 2.425
    82210Pb = -14.743 614C = 3.020

    For these 3 decays estimate the distance from the centre of the nucleus at which the Coulomb Barrier is equal to the calculated Q value (assuming e2/4[itex]\pi[/itex][itex]\epsilon[/itex]=1.44MeV.fm)

    2. Relevant equations

    Q = (Mintial-Mresultants)

    VC = Z1Z2e2/ 4[itex]\pi[/itex][itex]\epsilon[/itex]r


    3. The attempt at a solution

    For values of Q

    224Ra --> 220Rn + [itex]\alpha[/itex]

    Q = 18.818-10.604-2.425 = 5.8MeV

    224Ra --> 212Pb + 12C

    Q = 18.818+7.557-0 = 26.4MeV

    224Ra --> 210Pb + 14C

    Q = 18.818+14.743-3.020 = 30.5MeV

    And for the second part:

    VC = Z1Z2e2/ 4[itex]\pi[/itex][itex]\epsilon[/itex]r

    224Ra --> 220Rn + [itex]\alpha[/itex]

    5.8 = (88 x 82 x 1.44) /r
    r = 1791.5 fm

    I thought that this seemed quite a large value. Also, for the second 2 reactions I wasn't sure if this was the correct equation to be using because I thought it was specificaly just for alpha decay, and I didn't think the last 2 reactions were alpha.

    Any pointers would be fantastic!
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jan 22, 2012 #2
    It looks fine, and yes, that equation for Q-value should work for all nuclear decays.

    There is one problem though: you seemed to have added the product masses instead of subtracting them in the calculation of the Q-values for the second and third decays.
     
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