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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Sign up to join this communityThe 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}
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
$$ N'(t) = \frac{d}{dt} N_0 e^{-\lambda t} = - \lambda N_0 e^{-\lambda t} $$