# The law of radioactive decay: explanation of a formula

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: $$$$N'(t)=N_0(1-e^{-\lambda t})\, ?$$$$

• Your second equation shouldn't be the starting point of the derivation. How did you get there? – noah Mar 21 '19 at 22:12
• 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. – Sebastiano Mar 21 '19 at 22:16
• en.wikipedia.org/wiki/Radioactive_decay#One-decay_process . 5 secs of Googling. – Gert Mar 21 '19 at 22:17
• @Gert Could I please have a better explanation than Wikipedia, simpler and more complete? Actually, I haven't thought about searching on the web. – Sebastiano Mar 21 '19 at 22:20
• I don't think you'll find anything simpler (or more complete). It really is a very simple problem, you know? – Gert Mar 21 '19 at 22:23

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

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

• "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. – Gert Mar 21 '19 at 23:31
• oops.. made a typo thanks – InertialObserver Mar 21 '19 at 23:32
• @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. – Sebastiano Mar 22 '19 at 10:38