I recently learnt the derivation of radioactive decay formula and I am quite surprised about using integration to derive the formula.

But $N$ (the number of atoms) can only be discrete numbers (like 1,2 & not 1.5 or.9).
But won't using integration include all the continuous values of $N$?
So instead of using integration, shouldn’t some kind of summation be used to include the discrete values only?

This answer says something like this about the use of summation instead of integration. https://math.stackexchange.com/q/1509235/

$$ \begin{align} \frac{dN}{dt} & = - \lambda N \\ \end{align} $$ hence, $$ \begin{align} \int \frac{dN}{dt} & = - \lambda \int dt \\ \\ \ln{N} & = - \lambda t + C \end{align} $$ If the initial number of nuclei is $N_0$ and $N = N_0$ when $t = 0$
then (i becomes, $$ \begin{align} \ln N_0 = C \end{align} $$ substituting for C into (i $$ \begin{align} \ln{N} & = - \lambda t + \ln{N_0} \\ \ln{N} - \ln{N_0} & = - \lambda t \\ \ln({ \frac{N}{N_0} }) & = - \lambda t \\ \Rightarrow \frac{N}{N_0} & = e^{-\lambda t} \\ \Rightarrow N & = N_0 e^{- \lambda t} \end{align} $$

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    $\begingroup$ Please do not post images of texts you want to quote, but type it out instead so it is readable for all users and so that it can be indexed by search engines. For formulae, use MathJax instead. $\endgroup$
    – ACuriousMind
    Commented Mar 18, 2020 at 21:53
  • $\begingroup$ Additionally, your link to math.SE goes to a question about the generic difference between summation and integration, while your sentence talks about an answer that says "something like this" where it is unclear what "like this" refers to. Could you clarify why you feel this link is relevant to your question? $\endgroup$
    – ACuriousMind
    Commented Mar 18, 2020 at 21:57
  • $\begingroup$ @ACuriousMind The first answer in math.SE link says that Planck used summation instead of integration because the spectrum was built on discrete values. So shouldn't the same thing be applied for radioactive decay as the number of atoms is also discrete? $\endgroup$ Commented Mar 19, 2020 at 8:28
  • $\begingroup$ @ACuriousMind My skill in MathJax is poor. And so I just posted the screenshot. $\endgroup$ Commented Mar 21, 2020 at 7:57
  • $\begingroup$ @Theoretical That's okay for new users to do, but for people who are regular contributors here, we do expect them to learn how to use MathJax to express formulas, at least the straightforward ones. Don't worry about it too much now, but it would be really great if you could use MathJax instead of images of equations for future questions. You can always ask on Physics Meta or in chat for help if you need it. $\endgroup$
    – David Z
    Commented Mar 22, 2020 at 2:59

2 Answers 2


Since the atoms decay at random, $N$ should be properly understood as the expected number of atoms left: $$N = \sum_{n=0}^\infty n\, P(\text{there's $n$ atoms left})$$ Since probabilities are not integers, $N$ doesn't need to be an integer either, and it satisfies the continuous differential equation.

Alternatively you can use the differential equation to calculate the probability that any particular atoms has not yet decayed after time $t$ to get $$ P(\text{atom has not yet decayed}) = e^{-\lambda t}$$

The actual number of atoms left at any given point may differ from $N$ but for large amounts of atoms the difference will be relatively small.

  • $\begingroup$ Latosinski You are saying that $P$ can be fractional and so $N$ can also be a fraction. But won't $P$ be the precise values only for which $N$ and your summation will be integer? How can $N$ be a fraction just because $P$ isn't a fraction? $\endgroup$ Commented Mar 18, 2020 at 11:30
  • $\begingroup$ "But won't P be the precise values only for which N and your summation will be integer?" Sorry, I can't understand what you're asking about. Anyway, thre values of $P$ can be calculated from quantum mecanics, and it can actually be derived that $$P(\text{there's $n$ atoms left}) = \binom{N_0}{n} (e^{-\lambda t})^n (1-e^{-\lambda t})^{N_0-n}$$ It's a precise value (assuming that atoms decay independently at a constant rate). To show that $N$ doesn't have to be integer, consider $N_0=1$. What is expected value of the number of atoms left, if the probability that the atom decayed is $p$? $\endgroup$ Commented Mar 18, 2020 at 18:12
  • $\begingroup$ Latosinski Consider $N_0=5$ and $\lambda=1$ for example. Then after 2 second, $N$ will be $0.6766$. How does it make any sense? How can $N$ be a fraction? $\endgroup$ Commented Mar 20, 2020 at 3:43
  • $\begingroup$ @Theoretical $N$ can be a fraction because it's not an actual number of particles left, but the expected value,. And this doesn't mean "most probable" - it is the average number of particles you'd find if you were to repesat the experiment infinitely times. I'd like to matcch the answer to what you knwo, but for that you'd ned to tell me how much you know aboult probability theory. $\endgroup$ Commented Mar 20, 2020 at 9:29
  • $\begingroup$ @Theoretical In your case, $N_0=5$, $\lambda=2$, $t=1$, the precise analysis gives that there's approximately: 0.0045% chance there will be still 5 atoms left; 0.15% chance there will be 4 atoms left; 1.85% chance there will be 3 atoms left; 11.84% chance there will be 2 atoms left; 37.8% chance there will be 1 atom left; and 48.3% chance there will be 0 atoms left. If you sum all of these numbers with probabilities you get the expected number of atoms left to be approximately 0.6767. $\endgroup$ Commented Mar 20, 2020 at 9:40

If you restrict $N$ to be discreet, then you can also not write $dN/dt = -\lambda N$, as the derivative implies a continuous function of time. You should rather think of $N$ as the average number of atoms in an ensemble. We prepare a large number $M$ of identical setups, each containing $N_0$ atoms at $t=0$, and measure the average number of atoms in these ensembles at each point. If $M$ is large enough, we can treat $N$ approximately as a continuous function. (at the end, if you want, we can think of all the ensembles as one large setup, and it still works)


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