Poisson distribution

By 0x7df, Fri 18 September 2020, in category Uncategorized

mathematics, statistics

Consider a radioactive substance with concentration C, in atoms per litre, which is constant in space and time. The measured half-life is \(\tau\), in seconds, and the decay constant is \(\lambda = \ln(2)/\tau\). The number of decays per second per litre – the decay rate – is \(\lambda C\).

However, this is an average value, and the decay is a Poissonian process. The probability of \(k\) decays per litre in any given second-long period is:

$$ p(k) = \frac{\left[\lambda k\right]^k e^{-\left[\lambda C\right]}}{k!} $$

The figure shows example Poisson distributions for \(\lambda C\) = 10, 50, 90 and 130.

poisson_distributions

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