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. $A=\frac{dN}{dt}, A=\lambda N$. $\endgroup$ Commented 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. $\endgroup$ Commented 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
    Commented 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 $\delta t$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.