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Simple question about radioactivity equation

  1. Sep 27, 2011 #1
    In the radiactivity equation A = A0e-ln(2)t/T1/2 How do I get A0? Is that just ln(2)N0/T1/2? What if I don't know the initial number of atoms in the sample? Thanks...
     
  2. jcsd
  3. Sep 27, 2011 #2

    Astronuc

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    Staff: Mentor

    If one does not know No at to, one counts at A or N at t1 and t2, and then extrapolate back to to. One would also could also determine the relative amounts of decaying nuclide and daughter. Elements can be identified by chemical analysis, e.g., emission spectroscopy (perhaps with ICP) or mass spectrometry, and radionulides can be identified by characteristic radiation emissions. Usually one does a combination of analyses.
     
  4. Sep 27, 2011 #3
    hmm... Ok so here's the question from my book then.
    A sample X with Half-life 7.5min is measured from t1 = 3 min to t2=13 min. The total number of counts during those 10min is 34650. They want me to find the activity of the sample at t0=0... Any thoughts? Thanks for your help!
     
  5. Sep 27, 2011 #4

    Astronuc

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    If one is given the total counts between two times, then integrates the activity over time, i.e., between t1 and t2

    N = [itex]\int_{t_1}^{t_2} A(t) dt[/itex], and one should know the expression for A(t) = λ N(t), and one know the expression for N(t) related to No.
     
    Last edited: Sep 27, 2011
  6. Sep 27, 2011 #5
    Note that in general:

    Quantity = Rate X Time

    Shorthand,

    Q = R t

    or, in differential form:

    dQ = R dt

    And,

    Q = Intergral [R dt]

    On your case

    R = A(t)

    and you can find Ao.
     
  7. Sep 27, 2011 #6
    Ok, so the A(t2) = A(t1)*e-[itex]\lambda[/itex]t2 and solve for A(t1). But what is A(t2)? 34650/10min?
    Then A(t1) = A(t0)*e1[itex]\lambda[/itex]t1 and solve for A(t0) correct?
     
  8. Sep 27, 2011 #7

    Astronuc

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    Staff: Mentor

    No.

    One needs to work out the integral for the activity during the period from t1 to t2.
    The counts = 34650 represents all the decays during that period, which is found by integrating the activity A(t) between the two times. Work out the integral.

    Remember A(t) = λ N(t). But what is the expression for N(t)?
     
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