Radioactivity

Abstract

The main types of radioactivity are briefly described. This is followed by a section which explains the various different devices used for detecting and measuring radioactivity, with a short description of the workings of each device and a discussion of which devices are more suitable for the detection of the different types of radioactivity. The exponential decay of radioactive nuclides is explained and the mean lifetime and half-life are defined. Statistical fluctuations in the observed number of decay events due to the randomness of radioactivity are discussed, and the standard deviation is introduced as a measure of the error in a predicted value of the number of decay events. The application of radioactivity to radiometric and carbon dating is described. The chapter ends with brief discussions of multi-modal radioactive decay, decay chains and induced radioactivity.

Appendix 5: Poisson Distribution

The probability that a particular nucleus will undergo radioactive decay in a time interval Δt is λΔt, where λ is the decay constant . Likewise, the probability that it will not decay is $\left (1-\lambda \varDelta t\right )$.

If we have a sample of N nuclei and N  ≫ λΔt, then the probability, P(n), that exactly n nuclei will decay in time interval Δt is the probability that n nuclei will decay, multiplied by the probability that (N − n) nuclei will not decay, multiplied by ^N C _n, the number of ways of selecting n decaying nuclei out of a total of N nuclei.

$$\displaystyle \begin{aligned} P(n) \ = \ ^NC_n (\lambda \varDelta t)^n(1-\lambda \varDelta t)^{N-n}, {}\end{aligned} $$

(5.22)

$$\displaystyle \begin{aligned} \mbox{ where } ^NC_n \ = \ \frac{N!}{n!(N-n)!}. \end{aligned}$$

The average number of nuclei that will decay in this time interval is given by

$$\displaystyle \begin{aligned} \overline{n} \ = \ N \lambda \varDelta t, \end{aligned} $$

(5.23)

so we may rewrite (5.22) as

$$\displaystyle \begin{aligned} P(n) \ = \ \frac{N!}{n!(N-n)!} \left(\frac{\overline{n}}{N}\right)^n \left(1- \frac{\overline{n}}{N}\right)^{N-n}. \end{aligned} $$

(5.24)

P(n) is negligible except in the region $n \, \sim \, \overline {n}$, so we can always take n  ≪ N.

In this limit, we can make the standard approximations

$$\displaystyle \begin{aligned} \left(1- \frac{\overline{n}}{N}\right)^{N-n} \ \approx \ \left(1- \frac{\overline{n}}{N}\right)^{N} \ \approx \ e^{-\bar{n}}, \end{aligned}$$

and

$$\displaystyle \begin{aligned} (N-n)! \ \approx \ \frac{N!}{N^{n}}.\end{aligned}$$

Using this approximation P(n) reduces to the Poisson distribution:

$$\displaystyle \begin{aligned} P(n) \ = \ \frac{\overline{n}^n}{n!} e^{-\overline{n}}. \end{aligned} $$

(5.25)

We note that the total probability, $\sum _{n=0}^N P(n)$, is equal to 1, and that the average value,

$$\displaystyle \begin{aligned} \overline{n} \ = \sum_{n=0}^N n P(n), {} \end{aligned} $$

(5.26)

as expected.

We now calculate the average value of n(n − 1), using the relation

$$\displaystyle \begin{aligned} n(n-1) \overline{n}^n \ = \ \overline{n}^2 \frac{d^2}{d\overline{n}^2} \left(\overline{n}^n\right), \end{aligned}$$

so that

$$\displaystyle \begin{aligned} \overline{n(n-1)} \ = \sum_{n=0}^N n(n-1) P(n) \ = \ \overline{n}^2 e^{-\overline{n}} \sum_{n=0}^N \frac{d^2}{d\overline{n}^2} \left( \frac{\overline{n}^n}{n!} \right) \end{aligned} $$

(5.27)

For $N\,\gg \, \overline {n}$, we can take the limit N  → ∞ to obtain

$$\displaystyle \begin{aligned} \overline{n(n-1)} \ = \ \overline{n}^2 e^{-\overline{n}} \frac{d^2}{d\overline{n}^2} \left( e^{\overline{n}} \right) \ = \ \overline{n}^2. \ {} \end{aligned} $$

(5.28)

This gives us an expression for the mean of the square, $\overline {n^2}$:

$$\displaystyle \begin{aligned} \overline{n^2} \ = \ \overline{n}^2 + \overline{n}. {} \end{aligned} $$

(5.29)

The mean square deviation is

$$\displaystyle \begin{aligned} \overline{(n-\overline{n})^2} \ = \ \sum_{n=0}^N \left( n^2-2 n \overline{n} + \overline{n}^2\right) P(n) \ = \ \overline{n^2}-\overline{n}^2 {}, \end{aligned} $$

(5.30)

(where we have used (5.26)).

From (5.29) and (5.30), we see that the standard deviation is

$$\displaystyle \begin{aligned} \sigma \ = \ \sqrt{ \overline{(n-\overline{n})^2}} \ = \sqrt{\overline{n}}. \end{aligned} $$

(5.31)