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B Calculating activity of a radionuclide

  1. Mar 28, 2017 #1
    Activity (in Bq) of 1 mol of a radionuclide is given by formula:
    A = NA x ln(2) / t1/2 = 0.693 x NA / t1/2
    where NA is Avogadro number and t1/2 the half-life (in seconds)

    Why don't we use simply A = 0.5 x NA / t1/2 ?

    After all, t1/2 means that, after that time, half of atoms will have decayed...
     
  2. jcsd
  3. Mar 28, 2017 #2

    phyzguy

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    If I have N0 atoms in the sample, with a decay rate of λ, the number of atoms in the sample will decay as follows:
    [tex] N = N_0 e^{-\lambda t}[/tex]
    The activity is the number of decays per second, which is given by:
    [tex] A = -\frac{dN}{dt} = \lambda N_0 e^{-\lambda t} = \lambda N [/tex]
    The half-life is the time when 1/2 of the atoms have decayed, which from the first equation is given by:
    [tex] \frac{N}{N_0} = 1/2 = e^{-\lambda t_{1/2}} ;\,\,\, t_{1/2} = \frac{\log(2)}{\lambda} [/tex]
    So the activity is given by [itex] A = \frac{\log(2)}{t_{1/2}}N [/itex]. If we used "1/e-life" instead of "half-life", we wouldn't have this complication.
     
  4. Mar 29, 2017 #3
    So, calculating A = 0.5 x NA / t1/2 would give us the mean activity from t = 0 (present time) up to t = t1/2 ..., because during the length time t1/2, exactly 0.5 x NA atoms decay.

    But calculating A = NA x ln(2) / t1/2 will give the instant activity, when the number of atoms involved is the present number NA.

    In that case, OK, I understand the difference.
     
  5. Mar 29, 2017 #4

    phyzguy

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    Yes, what you said is correct.
     
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