# Finding radioactive nucleus given table of values

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

$$T_{1/2}=\frac{\ln2}{\lambda}=\frac{t\,\ln2}{\ln\left(N_{0}/N\right)}$$

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

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

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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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• Have you also calculated the associated errors of those half-lives? Commented May 16, 2013 at 8:21
• @UnkleRhaukus No, I don't. How can I calculate those errors? Commented May 16, 2013 at 15:20
• 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, Commented May 17, 2013 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:

• So, I suppose the corresponding isotope would be $^{131}I$, with $T_{1/2}=8.04\ d$. Commented May 16, 2013 at 5:22
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...