The plot in the interactive element above is drawn using time intervals that are much smaller than the half-life so a smooth curve is drawn. This has an impact on the probability of decay value that should be used. To see why, consider the following equation for how the number of nuclei changes over a time period: $$A_{t} = A_0 - pA_0 = A_0 \left( 1-P \; \right),$$ where \(P \;\) is the probability of decay over the time period. If the time period is one half-life then, by definition, \(P = 0.5\). Now consider taking two time steps, each of half a half-life. A first guess is that as the time-step is half as long, the probability of decay should be half as large, so that after the first half time-step the number of nuclei is $$A_{\frac{t}{2}} = A_0 \left( 1-\frac{P}{2} \; \right),$$ and after the second half time-step the final value is $$A_{t} = A_{\frac{t}{2}} \left( 1-\frac{P}{2} \; \right) = A_0 \left( 1-\frac{P}{2} \; \right) ^ 2.$$ Using \(P = 0.5\) in the above will give the wrong answer - it will not yield a reduction by half over the half-life. To achieve this a value of \(P = 0.586\) (3 d.p.) is required. This exercise can be extended to ever smaller sub-divisions of the half-life. When this is done the value of \(P \;\) converges to the natural logarithm of 2 (about 0.693).
The above showed that when the half-life interval is divided into two sub-intervals, the
number of nuclei is prediced by
$$A_{t} = A_0 \left( 1-\frac{P}{2} \; \right) ^ 2,$$
and as, by definition, \(A_{t} = \frac{1}{2}A_0\), a value of \(P \approx 0.586\) is required to satisfies this. The above can be extended, for example if the half-life interval
is divided into four sub-intervals, the number of nuclei is prediced by
$$A_{t} = A_0 \left( 1-\frac{P}{4} \; \right) ^ 4,$$
and generally when the interval is sub-divided into \(N\) substeps
$$A_{t} = A_0 \left( 1-\frac{P}{N} \; \right) ^ N,$$
and a value of \(P \;\) is required such that
$$\left( 1-\frac{P}{N} \; \right) ^ N = \frac{1}{2},$$
as \(N \rightarrow \infty\). To solve this, write
$$y = log_e \left( 1-\frac{P}{N} \; \right) ^ N = N log_e \left( 1-\frac{P}{N} \; \right)=\frac{log_e \left( 1-\frac{P}{N} \; \right)}{\frac{1}{N}}.$$
Allowing \(N \to \infty\) in the above yields \(\frac{0}{0}\). To avoid this
L'Hopital's rule is used:
$$\lim_{x\to c} \frac{f(x)}{g(x)}=\lim_{x\to c} \frac{f^\prime (x)}{g^\prime (x)}$$
Writing \(log_e \left( 1-\frac{P}{N} \; \right) = log_e \left( \frac{N-P}{N} \; \right)
= log_e \left( N-P\right) - log_e \left( N\right)\)
the top and bottom expressions of \(y\) are differentiated to give
$$\frac{f^\prime (x)}{g^\prime (x)} = \frac{\frac{P}{N(N-P)}}{-\frac{1}{N^2}} = \frac{-PN^2}{N^2-NP},$$
and
$$\lim_{N\to \infty} y = \lim_{N\to \infty}\frac{-PN^2}{N^2-NP} = -P.$$
Recall \( y = log_e \left( 1-\frac{P}{N} \; \right) ^ N\), and as it required that
\(\left( 1-\frac{P}{N} \; \right) ^ N = \frac{1}{2}\), it follows that in the limit
\(N\to \infty\)
$$-P = log_e{\frac{1}{2}} \Rightarrow P=log_e(2).$$
This result is as expected from a differential equation model of radioactive decay.
The process is modelled by
$$\frac{dA}{dt} = - \lambda A,$$
where \(\lambda\) is the rate of change constant. The solution to this is
$$A(t) = A_0 e^{-\lambda t}.$$
If the time period is one half life then when \(t = 1\), \(A = 0.5A_0\), so that
$$e^{-\lambda} = \frac{1}{2} \Rightarrow -\lambda = log_e(\frac{1}{2}) \Rightarrow \lambda = log_e(2).$$