# Radioactive decay: calculating probability of decay from half life/decay constant

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.

$$n_f=n_i2^{-\frac{t}{T_{1/2}}}\implies\frac{n_f}{n_i}=2^{-\frac{t}{T_{1/2}}}$$

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:

$$p(t_i,t_f)=\frac{\ln{2}}{T_{1/2}}\int_{t_i}^{t_f}2^{-\frac{t}{T_{1/2}}}dt$$

1. After further simplification which becomes:

$$p(t_i,t_f)=2^{-\frac{t_i}{T_{1/2}}}-2^{-\frac{t_f}{T_{1/2}}}$$

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

• 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. $A=\frac{dN}{dt}, A=\lambda N$. Commented Jul 5, 2023 at 22:19
• The probability of decay at exactly the 10th second is zero. The time duration is zero, not one second. Commented Jul 5, 2023 at 22:21
• @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. Commented Jul 6, 2023 at 16:46

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 $$\delta t$$.