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Error Propogation for Half-Life

  1. Dec 1, 2015 #1
    In this homework question we are told to calculate the half-life of an isotope based on count-rates before and after a given time interval, the relevant equation is given below.

    Half-life = t1/2 = tln(2)/ln (A/A0)

    The second part asks to determine the standard deviation of the half-life due to counting statistics. We have an uncertainty for the value A and the value A0.

    I have found the following solution but cannot quite follow it, so was hoping someone might be able to explain/walk through it with me.

    d/dt (t1/2) = − (ln (2) t/(A/A0))/ln((A/A0)2)

    Plugging in t = 24 and R = 2.875 yields 5.248 hr−1

    σ2t1/2 = (5.248)2 ((9.14/(41.4)2 + 16.83/(118.3)2) *(2.875)2

    Thus, we find that expected standard deviation of half-life is σ(t1/2) = 1.22 h

    Thanks in advance for any help.
     
  2. jcsd
  3. Dec 1, 2015 #2

    mfb

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    d/dt does not make sense at that point. If you calculate the time-derivative of the expression for the half-life, you get a different result.
    24 what, and what is R?
    The time-derivative of a time should be dimensionless.

    You can calculate the uncertainty on the ratio first, and then propagate this to the half-life measurement. ##-\frac{\ln( 2) t }{A/A_0 \ln(A/A_0)^2}## can be useful there as some part of a different formula.

    Can you give the numbers you use for your calculations?
     
  4. Dec 1, 2015 #3
    Thank you for the reply.

    t=24hr and R=A/A0=(41.4minute-1)/118.3minute-1

    I certainly agree with what you are saying about the time derivative of a time, I wasn't sure on this myself. The half-life itself is not a function of time anyways, so not sure how the time derivative makes sense or why it is invoked. It is possible that this is an error. I found this solution here and have been trying to see how it was arrived at. See the image of the solution below. The equation you have arrived at certainly is consistent with the first part of this solution and makes sense to me, however I am still unclear on how that carries over to the second part.
    upload_2015-12-1_15-26-29.png
    Many thanks for your help.

    Best Regards,
     

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  5. Dec 1, 2015 #4

    haruspex

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    If you look at the right hand side of the d/dt(t1/2) equation you can see that the derivative taken was with respect to R, not t.
     
  6. Dec 1, 2015 #5

    mfb

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    There are more issues. The ##\sigma S_i## calculations mix standard deviations (left and middle) and variance (right).
    ##\sigma R## should have been calculated in a clearer way, and so on. It is the left two brackets in the last formula.
    5.248 seems to be a bad approximation of the expression above, I get 5.188.
     
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