Question Measuring the number of decays per minute $N(t)$ of a radioactive source every four days we have that: $N(t=0):=N_0=200$, $N(t=4)=141$, $N(t=8)=100$, $N(t=12)=71$, where $t$ is measured in days ($d$). ¿Which radioactive isotope is it?

Answer From the decay law we have that $N(t)=N_{0}e^{-\lambda t}$, then $\lambda=\frac{1}{t}\ln\frac{N_{0}}{N}$, and


Inserting the values for $N(t=4)$ , I find that

$$T_{1/2}=7.93171\ d$$ Inserting the values for $N(t=8)$ , I find that

$$T_{1/2}=8.00000\ d$$ And inserting the values for $N(t=12)$ , I find that

$$T_{1/2}=8.03154\ d$$


My question is which one of these half-lives is the correct? And knowing the right value for $T_{1/2}$, how can I know the isotope?


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    $\begingroup$ Have you also calculated the associated errors of those half-lives? $\endgroup$ – Elements in Space May 16 '13 at 8:21
  • $\begingroup$ @UnkleRhaukus No, I don't. How can I calculate those errors? $\endgroup$ – Ana S. H. May 16 '13 at 15:20
  • $\begingroup$ Assuming you have already subtracted the background count, the distribution can be assumed to be Poisson, the probabilistic error in your counts measurements is equal to the square root of the number of counts, $\endgroup$ – Elements in Space May 17 '13 at 16:13

Since radioactivity is a random process, you'd expect some fluctuation in the number of decays, i.e. if you wait for the half life, there's no guarantee that exactly half of the nuclei will have decayed!

Based on your 3 estimates of the half life, you could just take the mean and go with that.

And how to find the isotope? There's lists for that. I just googled "half lives of radioactive isotopes" and, voila:


  • $\begingroup$ So, I suppose the corresponding isotope would be $^{131}I$, with $T_{1/2}=8.04\ d$. $\endgroup$ – Ana S. H. May 16 '13 at 5:22
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    $\begingroup$ Comes close enough for me :) $\endgroup$ – Lagerbaer May 16 '13 at 13:10

The "correct" half-life is the one that best fits all the experimental data available.

Start with your original equation,$$N(t)=N_{0}e^{-\lambda t}$$ Take natural logarithms of both sides and obtain:$$\ln (N(t))=\ln(N_0)-\lambda \ t$$Note that $N(0)$ is not the same as $N_0$. The first is an experimental point, with the same sort of random and experimental errors as the other three points have; the second is the parameter that, with some $\lambda$, best fits the data to the given equation.

Calculate the natural logarithms of each of the four count rates. Plot a graph of the four values of t and $\ln (N(t))$. The slope of the best straight line through these four points is $-\lambda$. Excel says it's 0.086263...

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    $\begingroup$ While this analysis is certainly better than the naive one, excel's fit neglects the counting error on each point. Domain specific tools such as ROOT have fitting routines that are more attuned to getting it really right. $\endgroup$ – dmckee May 16 '13 at 20:51

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