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}
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$\begingroup$ Your second equation shouldn't be the starting point of the derivation. How did you get there? $\endgroup$– noahMar 21, 2019 at 22:12
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$\begingroup$ 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. $\endgroup$– SebastianoMar 21, 2019 at 22:16
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$\begingroup$ en.wikipedia.org/wiki/Radioactive_decay#One-decay_process . 5 secs of Googling. $\endgroup$– GertMar 21, 2019 at 22:17
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$\begingroup$ @Gert Could I please have a better explanation than Wikipedia, simpler and more complete? Actually, I haven't thought about searching on the web. $\endgroup$– SebastianoMar 21, 2019 at 22:20
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$\begingroup$ I don't think you'll find anything simpler (or more complete). It really is a very simple problem, you know? $\endgroup$– GertMar 21, 2019 at 22:23
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1 Answer
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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
$$ N'(t) = \frac{d}{dt} N_0 e^{-\lambda t} = - \lambda N_0 e^{-\lambda t} $$
answered Mar 21, 2019 at 22:41
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$\begingroup$ "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. $\endgroup$– GertMar 21, 2019 at 23:31
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$\begingroup$ @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. $\endgroup$ Mar 22, 2019 at 10:38