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Nuclear Physics and Half-life

  1. Apr 22, 2007 #1
    1. The problem statement, all variables and given/known data

    A student takes a sample of 32,000 6He atoms and measures 31000 daughter nuclei 4 seconds later. What is the half-life of 6He? What is the daughter nucleus if the 6He underwent beta minus decay?

    2. Relevant equations

    [tex]N = (No)(0.5)^{t/halflife}[/tex]


    3. The attempt at a solution

    No = 32000 atoms = 32000 nuclei
    N = 31000 nuclei
    1 atom = 1 nuclei
    t = 4 s
    Find halflife

    [tex]N/No = (0.5)^{t/halflife}[/tex]


    [tex]log(N/No)/log(0.5) = t/halflife[/tex]

    [tex]log(N/No)/log(0.5) = t/halflife[/tex]

    [tex]halflife = t/log(N/No)/log(0.5)[/tex]

    [tex]halflife = t(log(0.5))/log(N/No)[/tex]

    [tex]halflife = (4 s)(log 0.5)/log(31000/32000)[/tex]

    I get half-life = 87.3 seconds, but the real half-life of 6He is about 806.7 ms. Does my answer and problem solving method look ok? Also, I think the daughter nucleus is 6Li because in beta minus decay causes the atomic number of the atom to go up by one as the neutron becomes a proton and an emitted electron. Does this make sense? I'm sorry for asking silly, easy questions, but we have not covered this material in class, so I am relying on my book to answer these questions. Thank you so much for your help.
     
  2. jcsd
  3. Apr 22, 2007 #2

    hage567

    User Avatar
    Homework Helper

    If you start with 32000 6He, and see 31000 daughter nuclei, what does that tell you about the change in the number of 6He nuclei? No is the # of 6He you started with, therefore N must be the # of 6He you have left after the 4 seconds. The number of 6He nuclei left is not 31000!
     
  4. Apr 22, 2007 #3
    Oh...now I understand, I hope.

    No = 32000 atoms = 32000 nuclei
    N = 32000-31000 = 1000 nuclei
    1 atom = 1 nuclei
    t = 4 s
    Find halflife

    [tex]N/No = (0.5)^{t/halflife}[/tex]


    [tex]log(N/No)/log(0.5) = t/halflife[/tex]

    [tex]log(N/No)/log(0.5) = t/halflife[/tex]

    [tex]halflife = t/log(N/No)/log(0.5)[/tex]

    [tex]halflife = t(log(0.5))/log(N/No)[/tex]

    [tex]halflife = (4 s)(log 0.5)/log(1000/32000)[/tex]

    Now this gives me half-life = 0.8 seconds, which is closer to the accepted value of 806.7 ms. I hope this is right. Thanks! :smile:
     
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