Let's say due to a nuclear reaction a radionuclide of half-life $T_{1/2}$ was created. I am trying to find out what will be the probability of that radionuclide undergoing radioactive decay within time $t_i$ and $t_f$ using the following method:

  1. I use the decay formula that relates time, initial and final numbers of particles and their half-life.


  1. I let $n_i \rightarrow\infty$ which turns $\frac{n_f}{n_i}$ into a probability distribution where $c$ is the normalizing constant.

$$p(t_i, t_f)=c\int_{t_i}^{t_f}2^{-\frac{t}{T_{1/2}}}dt\;\;\;\text{where,}\;\;c\int_{0}^{\infty}2^{-\frac{t}{T_{1/2}}}dt=1$$

  1. After doing the integration this is what the probability distribution becomes:


  1. After further simplification which becomes:


Did I get this right? If not please correct me.

  • $\begingroup$ Your math would be much easier if you use the decay constant ($\lambda$) rather than half-life. The average activity (decays per second) is proportional to the number of particles not yet decayed. $A=\frac{dN}{dt}, A=\lambda N$. $\endgroup$ Jul 5, 2023 at 22:19
  • $\begingroup$ The probability of decay at exactly the 10th second is zero. The time duration is zero, not one second. $\endgroup$ Jul 5, 2023 at 22:21
  • $\begingroup$ @KenMellendorf It was a source of my confusion as well, exactly at 10th second the probability of decay is 0, as we turned the decay equation into a probability distribution, but that just seemed a question too trivial to ask, as it can be answered even without doing any maths. That's why I supposed probably by 10th second the question actually meant between t=9s to t=10s. Thanks for clarifying. $\endgroup$
    – uran42
    Jul 6, 2023 at 16:46

1 Answer 1


You have the idea right, but the pedagogy is easier to follow if you think about the behavior of a population of radionuclides rather than about the probability distribution of a single decay. You can always go back to a single decay by asking what would happen if your total population were reduced to a single atom.

If you have a population $n_0$ at time $t=0$, then later on the population is

$$ n(t) = n_0 e^{-t/\tau} $$

which means the rate of change of the population is

$$ \frac{\mathrm dn(t)}{\mathrm dt} = -\frac{n_0}{\tau} e^{-t/\tau} $$

Here we're using the "lifetime" $\tau$, which is related to the half-life by $e^{-t/\tau} = 2^{-t/t_{1/2}}$, or $\tau\ln 2 = t_{1/2}$. Some people like the "decay constant" $\lambda = 1/\tau$, because numerators are less confusing that denominators. The half-life is great for explaining exponential decay to a non-mathematical audience, but $\tau$ and $\lambda$ are much more natural (pun intended) if you need to do calculus.

In the limit of an infinitesimal time interval, the probability of a decay is ${\mathrm dt}/{\tau} = \lambda\ \mathrm dt$.

If the time interval is comparable to the lifetime $\tau$, you're correct that you have to integrate the decays from your population over that interval. If we take your result and write $t_f = t_i + \delta t$, for the case where $t_{i,f}$ are both larger than $\tau$ but are close to each other, we can find

\begin{align} p(t_i, t_f) &= e^{-t_i/\tau} - e^{-t_f/\tau} \tag{your result} \\&= e^{-t_i/\tau}\cdot\left( 1-e^{-\delta t/\tau} \right) \\&\approx e^{-t_i/\tau} \cdot\frac{\delta t}{\tau} \end{align}

This is clearly just the probability that your nuclide has survived until $t_i$, multiplied by the probability that it decays during the brief interval $\delta t$.


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