Is the decay equation accurate at all?

Question

The decay equation is this for reference is $$N(t) = N(0) e^{-\lambda t} \, .$$

This suggest that any population of radioactive nuclei can never completely decay, which is not true, because some examples of isotopes have already decayed completely and doesn't exist

Furthermore, I will give an example, if there were a sample of 10^50 isotopes, and it takes one half life (however long that is) for the sample to be reduced to 10^49 isotopes, how is it possible that 2 of the same isotopes takes the same half life to be reduced to 1 isotope?

Answer 1

A half life is a statistical value.

First things first :

if there were a sample of $10^{50}$ isotopes, and it takes one half life (however long that is) for the sample to be reduced to $10^{49}$ isotopes,

This is wrong.

In one half life the amount is reduced by half not a tenth as you have here. That is the definition of half life - the time taken to reduce the quantity by one half.

how is it possible that 2 of the same isotopes takes the same half life to be reduced to 1 isotope ?

I'm going to assume you mean nuclei here, because isotope is a term for denoting nuclei with different numbers of neutrons but the same number of protons.

The approximation being used does not work on small numbers. It's designed to work only with large numbers. This is perfectly fine as most samples we need to work with will have enormous numbers of the isotope.

Applying it to small numbers is just pointless because e.g. neither of the two nuclei need decay ever. It's a completely random process and might never happen to an individual nucleus. It only makes sense to talk about half lifes on the scale of statistically large numbers of nuclei.

Answer 2

OK, if you want to be pedantic, $N(0)e^{-\lambda t}$ is the mean of $N(t)$, conditional on $N(0)$ having the specific value it's claimed to have, but no further information. After a time $t$, the number of surviving atoms is $\operatorname{B}(N(0),\,e^{-\lambda t})$-distributed, with $\frac{\sigma}{\mu}=\sqrt{\frac{e^{\lambda t}-1}{N_0}}$, so we can neglect the randomness in $N(t)$ if $N(0)\gg e^{\lambda t}-1$.

But the nuclei will almost surely all decay eventually (in fact, that italicized descriptor is also applicable for the fate of one specific nucleus). The above mean is deducible from each nucleus having a probability $e^{-\lambda t}$ of surviving at least a time $t$. This implies all nuclei decay in less than a time $t$ with probability $[1-e^{-\lambda t}]^{N(0)}$, and PDF $N(0)\lambda e^{-\lambda t}[1-e^{-\lambda t}]^{N(0)-1}$. The mean time until all nuceli decay is$$\begin{align}\int_0^\infty N(0)\lambda te^{-\lambda t}[1-e^{-\lambda t}]^{N(0)-1}dt&\stackrel{\star}{=}-\frac{1}{\lambda}\int_0^1N(0)x^{N(0)-1}\ln(1-x)dx\\&=-\frac{1}{\lambda}\left.\frac{d}{dc}\int_0^1N(0)x^{N(0)-1}(1-x)^cdx\right|_{c=0}\\&=\frac{\psi(N(0)+1)-\psi(1)}{\lambda}\\&=\frac{H_{N(0)}}{\lambda},\end{align}$$where $\stackrel{\star}{=}$ uses $x=1-e^{-\lambda t}$. See these links if you want to estimate the above mean with greater precision than $\frac{\ln N(0)}{\lambda}$, which is an unsurprising first-order timescale.