I have done a laboratory session at my university where I had to check that the disintegration of nucleii follows a Poisson distribution

$$P(n)=\frac{\lambda^n}{n!} e^{-\lambda},$$

where $P(n)$ is the probability of the disintegration of exactly $n$ nuclei for a given time interval.

I measured the data using a computer program and I've been analyzing the data using MatLab but I don't know if the results that I get make sense or not.

The activity of the test sample was $3$ disintegrations per second and when analyzing the data that I got manually I get that it fits via a Poisson distribution of $\lambda=3.45$ but when using the data that I got via the computer I get $\lambda \approx 900$. I believe that this is an error in my calculations.

This leaded me to wonder what is the exact meaning of $\lambda$ in that expression. I understand that it is the mean number of nuclei disintegrated in a given time, but, does that mean that it depends on the size of the sample? What is the meaning of this parameter?


If you have a Poisson distribution with a probability $$P(n)=\frac{\lambda^n}{n!}e^{-\lambda}$$ that there will be $n$ events per bin, then $\lambda$ is the mean number of events per bin. You can get this via a direct calculation, $$ ⟨n⟩ =\sum_{n=0}^\infty nP(n) =\sum_{n=0}^\infty n \frac{\lambda^n}{n!}e^{-\lambda} =\lambda e^{-\lambda}\sum_{n=1}^\infty \frac{\lambda^{n-1}}{(n-1)!} =\lambda, $$ or if you're deriving the Poisson form for $P(n)$ in your problem then you'll start with essentially that understanding for $\lambda$ and work your way up.

For your case, the mean rate of events will be related to the half-life of the sample but it will also be proportional to the length of the time bins and the amount of radioactive material, as well as affected by e.g. the detection efficiency of your apparatus. As to why your calculation gives a higher value than expected - that's impossible to answer without more details.

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