# Do unstable nuclei affect each other for decaying?

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.

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.

• Are you familiar with the coin-tossing demo of half-life? physics.stackexchange.com/a/633564/123208 Commented Apr 26, 2022 at 8:05
• @PM2Ring, that's exactly my problem. A single student who doesn't care about class and other students, won't have a watch and won't toss coins many times. Commented Apr 26, 2022 at 8:11
• An independent student, leaves the classroom any moment he felt so. Commented Apr 26, 2022 at 8:12
• Ok, but radioactive atoms don't behave like that. They have a characteristic mean lifetime, so they behave like the good students with the watches. Commented Apr 26, 2022 at 8:14
• No, it's not collective, but it is global. Each student, in each minute, follows the global rule, but he doesn't look at what the other students are doing, he just looks at his own watch & coin. The watches don't need to be synchronised (that would be collective), but they do need to tick at the same rate (in order to follow the global rule). Commented Apr 26, 2022 at 8:18

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 $$t=0$$.

• can we deduce from the sentence the rate of nuclear decay is constant that the decay is not a random phenomenon? Can we rephrase educational claims to *nuclear decay is an intrinsic property of the nucleus and is dependent upon the element and is not random"? Commented Apr 26, 2022 at 8:32
• Because I can't wrap my head around that randomness and see that curved line for randomness. I can't think of watch and coin analogy and see randomness in it. Repeated randomness might be a better word. That these nuclei are aware of time somehow and in each period (like in each orbit of wave-electrons around the nucleus) they try to decay once. Commented Apr 26, 2022 at 8:32
• @SaeedNeamati Q1. No we cannot as explained in the edit to my answer. Q2. Nuclear decay is random. Commented Apr 26, 2022 at 8:44
• @SaeedNeamati Mathematics sometimes leads to unexpected but indisputable conclusions, which makes it fun. Commented Apr 26, 2022 at 8:45

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.

• Can you show me a diagram of a system of dice, without a watch and without an interval? Can you show me if the graph is linear or curved? Commented Apr 26, 2022 at 16:16

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.

• Nope. A nucleus does not know about time interval. Time interval and half-life is the names we assign to a phenomenon. A single independent nucleus sits there and knows nothing about time interval and the probability makes no sense for it. It either decays at any moment, or it doesn't. Your argument is based on human perception of a population-based phenomenon. Think about it from the perspective of one single nucleus. Commented Apr 26, 2022 at 7:29
• In fact, your argument corroborates the dependence upon population. As if a nucleus is there and knows that first round is finished and now it is in the second round. Commented Apr 26, 2022 at 7:31
• @SaeedNeamati No, that's just your misconception about how probabilities adds up. If nucleus total decay probability would be dependent on previous rounds somehow,- then probabilities would add up as $$P(\Delta t_1 \cap \Delta t_2)=P(\Delta t_1) \cdot P(\Delta t_2|\Delta t_1)$$. Now atom knows nothing about previous non-decaying outcomes, thus total decay probability is $$P(\Delta t_1 \cap \Delta t_2)=P(\Delta t_1) \cdot P(\Delta t_2)$$. So get better prepared at probabilities, there's a lot of them in quantum mechanics, get used to them. Commented Apr 26, 2022 at 7:51
• Yeah, you're right. Thank you. However, I still don't understand the randomness nature of this process. For me based on my knowledge, periodicity (time-interval) makes no sense in randomness. Any nucleus either decays once or does not decay. Or otherwise, this is not random and it has an internal clock. Commented Apr 26, 2022 at 8:41
• "Having a clock" doesn't mean that some event has to be pre-programmed to fire at specific time. Consider another example with bacterias, which doesn't divide anymore (for example too old, etc.). Let's assume that bacteria will live 10 mins on average. Each time you'll look at the sample - you'll see that some bacterias has died. You don't know which bacteria will be dead next, nor when some bacteria will die exactly. But if you check sample at constant intervals, you'll notice that dead bacteria population increases exponentially as duration approaches 10 minutes. Commented Apr 26, 2022 at 20:24