The law of radioactive decay can be expressed in terms of $\,\tau=1/\lambda$ (average life) as:

$$ N(t)=N_0e^{-t/\tau}, \quad \tag{1} $$

Why deriving the (1) I have: \begin{equation} N'(t)=N_0(1-e^{-\lambda t})\, ? \end{equation}

  • $\begingroup$ Your second equation shouldn't be the starting point of the derivation. How did you get there? $\endgroup$
    – noah
    Commented Mar 21, 2019 at 22:12
  • $\begingroup$ I have some notes of a research that I'm elaborating. I have finded this without any linkage. I don't get it. If I derive the (1) I don't get the second one. $\endgroup$
    – Sebastiano
    Commented Mar 21, 2019 at 22:16
  • $\begingroup$ en.wikipedia.org/wiki/Radioactive_decay#One-decay_process . 5 secs of Googling. $\endgroup$
    – Gert
    Commented Mar 21, 2019 at 22:17
  • $\begingroup$ @Gert Could I please have a better explanation than Wikipedia, simpler and more complete? Actually, I haven't thought about searching on the web. $\endgroup$
    – Sebastiano
    Commented Mar 21, 2019 at 22:20
  • $\begingroup$ I don't think you'll find anything simpler (or more complete). It really is a very simple problem, you know? $\endgroup$
    – Gert
    Commented Mar 21, 2019 at 22:23

1 Answer 1


It comes from solving the differential equation

$$\frac{dN}{dt} = -\lambda N(t). $$

This equation comes from observations of the number of decay events $N(t)$. It's found through experiment that the rate of decay over a given time interval is proportional to the number of events recorded during that time. You can arrive at this conclusion by plotting the rate vs the number of events on a log log plot and finding that it is linear.

Formally, this is a differential equation. But solving it is really just a fact which you know already.

Which function $N(t)$ can you take the derivative of and get itself back times a constant?

The answer is exponentials, and so the solution to this equation is

$$ N(t) = N(0) e^{-\lambda t}. $$

Edit: I should also note that you took the derivative incorrectly. The correct derivative is

$$ N'(t) = \frac{d}{dt} N_0 e^{-\lambda t} = - \lambda N_0 e^{-\lambda t} $$

  • $\begingroup$ "the rate of decay is proportional to the the rate of these events" That doesn't make a whole lot of sense. By your own formula (which is correct) the decay rate is proportional to the number of remaining atoms, macroscopically speaking at least. $\endgroup$
    – Gert
    Commented Mar 21, 2019 at 23:31
  • $\begingroup$ oops.. made a typo thanks $\endgroup$ Commented Mar 21, 2019 at 23:32
  • $\begingroup$ @InertialObserver The derivative that I was founded is the same as your :-). My notes are probably wrong. In fact from the Wikipedia link provided by Gert: $N_{B}=N_{A_0}-N_A=N_{A_0}(1-e^{-\lambda t})$. Thank you very much for your answer. $\endgroup$
    – Sebastiano
    Commented Mar 22, 2019 at 10:38

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.