Do unstable nuclei affect each other for decaying?

Question

I am thinking about beta decay.

If we graph decayed nuclei count over time, we don't see a linear line. Rather, it would be a curved line.

I imagine myself as an unstable nucleus.

If I don't care about othe nuclei, I should decay randomly.

But if I see my surroundings, and somehow be under the influence of our population, then as we get fewer in count, I become lazier to decay, and as we increase in number, others somehow affect me to decay sooner.

I can't find an answer for why decaying graph is not linear, if nuclei decay independently.

I know about this question and could not get my answer:

Why does the same proportion of a radioactive substance decay per time period? (half life)

Update

The watch & coin analogy, while proving the independence of nuclei from each other, creates another problem.

This means that a nucleus has an internal periodic clock/mechanism. And in each period, it tries to decay once.

As an example, there might be a periodic behavior in quarks and gluons, that kick in based on the number of nucleons per nucleus (hence different half-lives), and when it happens nucleus either breaks or tolerates the change for the next cycle.

In other words, watch & clock analogy shows that decay is not a random process. It's a phenomenon that can be discovered and formulated.

Answer 1

The rate of nuclear decay is constant, say $-\alpha$. [Note added after comment: the probability of decay is constant in time, not that the actual number of decays]. However, the number of nuclei that have not decayed decreases. This leads to an exponential decay law as $$dn/dt = -\alpha n$$ has the solution $$n = n_0 e^{-\alpha t} \,,$$ where $n_0$ is the number of, all undecayed, nuclei at $t=0$.

Answer 2

This is a classic fallacy of probability known as the "gambler's fallacy". Basically, the idea is that it "feels" very strongly like that, if a dice roll comes up with ones 3 times in a row, then it is "unlikely" now - less likely than any of the other times, that it will come up a one the next time. Yet, so long as the die truly is a fair die, then that die will still have exactly the same probability to return a 3 this time as it did the previous 3 times.

Likewise, in your case, you seem to be imagining that as the nuclei around the one decay, that somehow that implies that its time is "coming due" and it's got to decay pretty soon. But this is just the same thing: it's psychology, it has nothing to do with the actual probability.

Answer 3

Nuclear decay law can't be a linear one. Let's say that a probability that particular nuclei has not decayed in time interval $\Delta t_x$ is $P_{\Delta t_x} = 0.9$. So probability that particular nuclei has not decayed in initial time interval $\Delta t_1$ is $P_{\Delta t_1} = 0.9$, in time interval $\Delta t_2 $ $\to$ $P_{\Delta t_2} = 0.9$, in time interval $\Delta t_3 \to$ $P_{\Delta t_3} = 0.9$, and so on, for the same length time intervals. Now ask what will be probability that nuclei will not decay between time intervals $t_n - t_1 = n\Delta t_x$, what's the probability $P_{n\Delta t_x} =~?$ Decay is fully random, particular nuclei is not affected by any other decaying nuclei, so "staying alive" probabilities will add in independent way, i.e.: probability that nuclei will not be decayed in 3 multiple time intervals is: $$ P_{3\Delta t_x} = P_{\Delta t_1}\cdot P_{\Delta t_2} \cdot P_{\Delta t_3} = 0.9^3 = 0.7$$ And in general probability of not being decayed over n-th time intervals will be : $$ P_{n\Delta t_x} = P_{\Delta t_1}\cdot P_{\Delta t_2} \cdot P_{\Delta t_3}\cdot ~\ldots~ \cdot P_{\Delta t_n} = \prod_{i=1}^n P_{\Delta t_i}$$ So even at first probability that particular atom has not been decayed is high ($0.9$), after just 50 identical time intervals it will become $P_{50\Delta t_x} = 0.005$,i.e. very low. That's the root cause why we get exponential decay law.